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Physics · Secondary School · Thermal Physics

Heat and Temperature

Coffee cooling on the table, the gaps left between railway tracks, ice melting in a glass and a tyre that feels fuller in the afternoon are all cases of thermal energy on the move, and you have probably met most of them without giving them a second thought. We want to understand what temperature really measures, how heat flows from one body to another and why it sometimes does not change the temperature at all.

  1. 1Temperature and particle motion
  2. 2Temperature scales
  3. 3Heat and thermal equilibrium
  4. 4Mechanical equivalent of heat
  5. 5Thermal expansion
  6. 6Heat capacity and specific heat
  7. 7Sensible heat
  8. 8Phase changes
  9. 9Calorimeter
  10. 10Heat transfer
  11. 11Ideal gases
  12. ✓Challenges
STEP 1

What does a thermometer really measure?

Everything is made of particles (atoms and molecules) that are always moving, and temperature measures how strongly these particles jiggle about. The hotter something is, the faster they move.

More precisely, absolute temperature (in kelvin) is proportional to the average kinetic energy of the particles. At absolute zero (0 K, or −273 °C) the motion is the smallest possible, and nothing can be colder than that.

The particles do not all have the same speed, because they collide all the time and swap energy, one coming out of each collision faster and the other slower. What we get in the end is always the same distribution, discovered by Maxwell, with a few very slow particles, a few very fast ones and most of them near a most probable speed.

\(\overline{E_k} \propto T\) (in kelvin)\(\bar v \propto \sqrt{T}\)\(f(v) \propto v^2\, e^{-m v^2/2kT}\)Doubling the absolute temperature doubles the average energy, while the speed grows by only √2 ≈ 1.41 times. The simulation is in 2D, where the Maxwell distribution is \(f(v) \propto v\,e^{-mv^2/2kT}\), whereas in 3D space the factor is \(v^2\). In air at 27 °C, the most probable speed of a nitrogen molecule is about 420 m/s.

Let's discuss

  • If we bring the temperature down to 0 K, what happens to the particles?
  • Going from 300 K to 600 K, our first guess may be that the average speed doubles. Check the number in the readings.
  • Click ‘All at the same speed’ and watch what the collisions do to the histogram over the next few seconds.
  • Heat from 300 K to 600 K. Does the Maxwell curve change shape, or does it just stretch?
  • A metal door handle and a wooden door are at the same temperature, and yet the metal ‘feels’ colder. What could explain that?
  • Are temperature and heat the same thing? (We will answer this in step 3.)
In kelvin
In Celsius
Average energy (vs. 300 K)
Average speed (vs. 300 K)
Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    On a cold morning you touch a metal door handle and a wooden door, both at 15 °C. The metal feels much colder. Are they really at different temperatures?

    Show solution
    Both have been in the same surroundings for a long time, so they are at the same temperature, 15 °C.
    Your hand (≈ 36 °C) loses heat to both, but metal is a good conductor and pulls heat out of the skin much faster.
    What we feel as ‘cold’ is how quickly the skin loses heat, which is a different thing from the temperature of the object.
    No: the temperature is the same; the metal simply conducts heat faster than the wood.
  2. basic

    What does absolute zero (\(0\ \text{K} \approx -273\ ^\circ\text{C}\)) mean? Can a body be at \(-10\ \text{K}\)?

    Show solution
    Temperature measures how strongly the particles are moving.
    At absolute zero the thermal motion is the smallest possible, and no more motion energy can be taken from the particles.
    That is why there is no temperature below \(0\ \text{K}\), and the Kelvin scale has no negative values.
    \(0\ \text{K}\) is the minimum motion; \(-10\ \text{K}\) is impossible.
  3. basic

    A classmate says that ‘this cup of tea has a lot of heat’. From a physics point of view, what is wrong with that sentence, and how would you fix it?

    Show solution
    Temperature is a property of a body, and it measures the average motion of its particles.
    Heat is energy in transit, flowing from a hotter body to a colder one, so a body does not ‘have’ heat.
    The correct statement would talk about the temperature of the tea, or say that the tea can give off heat to the surroundings.
    Correction: ‘the tea is at a high temperature’ (heat is energy being transferred, not something stored).
  4. basic

    The average kinetic energy of the particles of a gas is proportional to its absolute temperature. If the gas goes from 200 K to 400 K, what happens to this energy?

    Show solution
    \(E_k \propto T\) (in kelvin).
    The ratio of the temperatures is \(\dfrac{400}{200} = 2\).
    So the average kinetic energy is also multiplied by 2.
    The average kinetic energy doubles.
  5. intermediate

    A gas at 27 °C is heated to 327 °C. By what factor is the average kinetic energy of the particles multiplied? Use \(T(\text{K}) = T(^\circ\text{C}) + 273\).

    Show solution
    In kelvin, the temperatures are \(27 + 273 = 300\ \text{K}\) and \(327 + 273 = 600\ \text{K}\).
    \(E_k \propto T\), so the factor is \(\dfrac{600}{300} = 2\).
    It would be wrong to compute \(\dfrac{327}{27} \approx 12\), because the proportion only holds on the absolute scale.
    The average kinetic energy doubles (factor 2).
  6. intermediate

    The average speed of the particles of a gas grows with \(\sqrt{T}\) (\(T\) in kelvin). If the temperature goes from 300 K to 1200 K, by what factor is the average speed multiplied?

    Show solution
    The ratio of the temperatures is \(\dfrac{1200}{300} = 4\).
    Since \(v \propto \sqrt{T}\), the speed factor is \(\sqrt{4} = 2\).
    The average speed doubles (factor 2).
  7. intermediate

    In a mercury thermometer, the column is 4 cm long in melting ice (0 °C) and 24 cm in boiling water (100 °C). The column expands uniformly. What is the temperature when it reads 13 cm?

    Show solution
    The thermometer works because the liquid expands in proportion to the change in temperature.
    From 0 °C to 100 °C the column grows \(24 - 4 = 20\ \text{cm}\).
    At 13 cm it has grown \(13 - 4 = 9\ \text{cm}\) above the ice mark.
    \(T = \dfrac{9}{20} \cdot 100 = 45\ ^\circ\text{C}\).
    \(T = 45\ ^\circ\text{C}\)
  8. intermediate

    Compare a gas at −73 °C with the same gas at 127 °C. What is the ratio of the average kinetic energies of the particles (cold / hot)? Use \(T(\text{K}) = T(^\circ\text{C}) + 273\).

    Show solution
    \(-73 + 273 = 200\ \text{K}\) and \(127 + 273 = 400\ \text{K}\).
    \(E_k \propto T\), so \(\dfrac{E_{\text{cold}}}{E_{\text{hot}}} = \dfrac{200}{400} = 0.5\).
    The average kinetic energy of the cold gas is half as much (ratio 0.5).
  9. challenge

    The average speed of the molecules of a gas at 27 °C is 500 m/s. What will this speed be at 927 °C? Use \(v \propto \sqrt{T}\) and \(T(\text{K}) = T(^\circ\text{C}) + 273\).

    Show solution
    \(T_1 = 27 + 273 = 300\ \text{K}\); \(T_2 = 927 + 273 = 1200\ \text{K}\).
    \(\dfrac{T_2}{T_1} = \dfrac{1200}{300} = 4\).
    \(v_2 = v_1 \cdot \sqrt{4} = 500 \cdot 2 = 1000\ \text{m/s}\).
    \(v \approx 1000\ \text{m/s}\)
  10. challenge

    The average kinetic energy of a gas molecule is \(E = \dfrac{3}{2}\,k\,T\), with \(k = 1.38 \cdot 10^{-23}\ \text{J/K}\). Calculate \(E\) at 300 K and find the temperature, in °C, the gas must reach for \(E\) to triple. Use \(T(\text{K}) = T(^\circ\text{C}) + 273\).

    Show solution
    \(E = 1.5 \cdot 1.38 \cdot 10^{-23} \cdot 300 \approx 6.21 \cdot 10^{-21}\ \text{J}\).
    For \(E\) to triple, \(T\) must also triple, which gives \(3 \cdot 300 = 900\ \text{K}\).
    In Celsius, this is \(900 - 273 = 627\ ^\circ\text{C}\).
    \(E \approx 6.21 \cdot 10^{-21}\ \text{J}\); the gas must reach \(900\ \text{K} = 627\ ^\circ\text{C}\).
STEP 2

One temperature, three numbers

A temperature scale picks two fixed points and divides the interval between them. On the Celsius scale ice melts at 0 and water boils at 100, while on the Fahrenheit scale the same points are 32 and 212. The Kelvin scale starts at absolute zero and uses steps the same size as a Celsius degree.

Since the three thermometers show the same height, the fraction of the way between ice and steam is the same on all of them, and that is where the conversion formula comes from.

\(\dfrac{C}{5} = \dfrac{F - 32}{9} = \dfrac{K - 273}{5}\)\(\Delta C = \Delta K\)\(\Delta F = 1.8\,\Delta C\)Ice: 0 °C = 32 °F = 273 K · Steam: 100 °C = 212 °F = 373 K · Kelvin does not use ‘degree’, so we say 300 K and not 300 °K.

Let's discuss

  • Click ‘−40’ and look at the three readings. What is special about this temperature?
  • A change of 10 °C corresponds to how many kelvin, and to how many °F?
  • Can a temperature in kelvin be negative? Why?
  • Invent your own scale, with ice at 10 °X and steam at 90 °X. What does 50 °C read on it?
Celsius
Fahrenheit
Kelvin
Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    Normal human body temperature is about 37 °C. What is that in kelvin? Use \(K = C + 273\).

    Show solution
    \(K = C + 273\)
    \(K = 37 + 273 = 310\)
    \(310\ \text{K}\)
  2. basic

    An American tourist sees on their phone that it is 25 °C in Salvador, Brazil. What is that in °F? Use \(F = 1.8\,C + 32\).

    Show solution
    \(F = 1.8 \cdot 25 + 32\)
    \(F = 45 + 32 = 77\)
    \(77\ ^\circ\text{F}\)
  3. basic

    Why are there no negative temperatures on the Kelvin scale, while on Celsius and Fahrenheit they are common?

    Show solution
    Celsius and Fahrenheit put their zero at points chosen for convenience (melting ice, a laboratory mixture).
    The zero of the Kelvin scale is absolute zero, the minimum motion of the particles, and there is nothing below it.
    Kelvin starts at absolute zero, so it only has values \(\geq 0\).
  4. basic

    An American thermometer reads 102.2 °F for a child. What is that in °C? Does the child have a fever? Use \(\dfrac{C}{5} = \dfrac{F - 32}{9}\).

    Show solution
    \(C = \dfrac{5\,(F - 32)}{9}\)
    \(F - 32 = 102.2 - 32 = 70.2\)
    \(C = \dfrac{5 \cdot 70.2}{9} = 39\)
    \(39\ ^\circ\text{C}\) is above \(37.5\ ^\circ\text{C}\), so it is a fever.
    \(39\ ^\circ\text{C}\) — yes, it is a fever.
  5. intermediate

    The average surface temperature on Mars is about −63 °C. Express it in kelvin and in Fahrenheit. Use \(K = C + 273\) and \(F = 1.8\,C + 32\).

    Show solution
    \(K = -63 + 273 = 210\ \text{K}\)
    \(F = 1.8 \cdot (-63) + 32\) \(= -113.4 + 32\) \(= -81.4\ ^\circ\text{F}\)
    \(210\ \text{K}\) and \(-81.4\ ^\circ\text{F}\)
  6. intermediate

    One afternoon, the temperature rose from 15 °C to 35 °C. What was the change in kelvin and in Fahrenheit? Use \(\Delta K = \Delta C\) and \(\Delta F = 1.8\,\Delta C\).

    Show solution
    \(\Delta C = 35 - 15 = 20\ ^\circ\text{C}\)
    \(\Delta K = \Delta C = 20\ \text{K}\) (the divisions are the same size)
    \(\Delta F = 1.8 \cdot 20 = 36\ ^\circ\text{F}\)
    \(\Delta K = 20\ \text{K}\) and \(\Delta F = 36\ ^\circ\text{F}\)
  7. intermediate

    There is one temperature at which Celsius and Fahrenheit thermometers show the same number. Which is it? Use \(\dfrac{C}{5} = \dfrac{F - 32}{9}\).

    Show solution
    Set \(C = F = x\).
    \(\dfrac{x}{5} = \dfrac{x - 32}{9}\) \(\Rightarrow 9x = 5x - 160\)
    \(4x = -160\) \(\Rightarrow x = -40\)
    \(-40\ ^\circ\text{C} = -40\ ^\circ\text{F}\)
  8. intermediate

    Between melting ice and boiling water there are 100 divisions on the Celsius scale and 180 on the Fahrenheit scale. Use this to explain why \(\Delta F = 1.8\,\Delta C\).

    Show solution
    Ice: \(0\ ^\circ\text{C} = 32\ ^\circ\text{F}\); steam: \(100\ ^\circ\text{C} = 212\ ^\circ\text{F}\).
    The same physical interval is worth \(100\ ^\circ\text{C}\) or \(212 - 32 = 180\ ^\circ\text{F}\).
    Each Celsius degree is equivalent to \(\dfrac{180}{100} = 1.8\) Fahrenheit degrees.
    A change of \(1\ ^\circ\text{C}\) equals a change of \(1.8\ ^\circ\text{F}\), so \(\Delta F = 1.8\,\Delta C\).
  9. challenge

    You create scale X: melting ice reads 10 °X and boiling water reads 90 °X. What does scale X read at 40 °C? And at what temperature do the X and Celsius scales show the same value?

    Show solution
    The relation between the two scales is \(\dfrac{X - 10}{90 - 10} = \dfrac{C - 0}{100 - 0}\) \(\Rightarrow X = 10 + 0.8\,C\).
    For \(C = 40\), we get \(X = 10 + 0.8 \cdot 40 = 10 + 32 = 42\ ^\circ\text{X}\).
    For the same value on both scales, \(C = 10 + 0.8\,C\) \(\Rightarrow 0.2\,C = 10\) \(\Rightarrow C = 50\).
    \(40\ ^\circ\text{C} = 42\ ^\circ\text{X}\); the scales coincide at 50 (\(50\ ^\circ\text{C} = 50\ ^\circ\text{X}\)).
  10. challenge

    At what temperature is the Fahrenheit reading exactly twice the Celsius reading? Use \(F = 1.8\,C + 32\).

    Show solution
    The condition is \(F = 2\,C\).
    \(2\,C = 1.8\,C + 32\) \(\Rightarrow 0.2\,C = 32\) \(\Rightarrow C = 160\).
    \(F = 2 \cdot 160 = 320\) (check: \(1.8 \cdot 160 + 32 = 320\)).
    \(160\ ^\circ\text{C} = 320\ ^\circ\text{F}\)
STEP 3

Heat: energy that flows from hot to cold

Heat is energy in transit, flowing on its own from the hotter body to the colder one, rather than something a body ‘has’. The flow stops when the temperatures become equal, and we call this state thermal equilibrium.

The zeroth law says that if A is in equilibrium with B, and B with C, then A is in equilibrium with C. A thermometer works because of this law, since it reaches equilibrium with the body and then shows that body's temperature.

The body whose temperature changes more is the one with the smaller heat capacity C = m·c, which we can think of as the one that is ‘easier to heat’.

\(T_{eq} = \dfrac{C_A\,T_A + C_B\,T_B}{C_A + C_B}\)\(C = m\,c\)This holds only if neither body changes phase and no heat is lost to the outside.

Let's discuss

  • With equal masses of water, where does the final temperature end up, and where does it go when B is much bigger?
  • Switch B to iron. Why does the iron change temperature so much, while the water changes so little?
  • Is the heat lost by A equal to the heat gained by B? Compare it with the ‘Heat transferred’ reading.
  • Why does a clinical thermometer need to stay on the body for a few minutes?
T of A
T of B
Predicted Teq
Heat transferred
Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A body A is in thermal equilibrium with a thermometer, and a body B is also in equilibrium with the same thermometer. What can you say about A and B? Which law is this?

    Show solution
    Thermal equilibrium means the same temperature.
    A has the thermometer's temperature, and so does B; therefore A and B have the same temperature.
    If they are put in contact, no heat will flow between them.
    A and B are in thermal equilibrium with each other (zeroth law of thermodynamics).
  2. basic

    A hot coffee is left on the table in a room at 25 °C. Where does the heat go, and until when does the coffee cool down?

    Show solution
    Heat always flows from the hotter body to the colder one, here from the coffee to the air and the table.
    The flow gets smaller as the temperatures get closer.
    It stops when the coffee reaches thermal equilibrium with the surroundings.
    Heat goes from the coffee to the surroundings, until the coffee reaches 25 °C.
  3. basic

    What is the heat capacity of 200 g of water? Use \(C = m\,c\) and \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(C = m \cdot c\)
    \(C = 200 \cdot 1 = 200\ \text{cal}/{}^\circ\text{C}\)
    This means 200 cal are needed to raise this water by 1 °C.
    \(C = 200\ \text{cal}/{}^\circ\text{C}\)
  4. basic

    How much heat must 500 g of water absorb to go from 20 °C to 30 °C? Give it in calories and in joules. Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(1\ \text{cal} \approx 4.2\ \text{J}\).

    Show solution
    \(Q = m \cdot c \cdot \Delta T\)
    \(Q = 500 \cdot 1 \cdot (30 - 20) = 5000\ \text{cal}\)
    In joules, this is \(5000 \cdot 4.2 = 21\,000\ \text{J}\)
    \(Q = 5000\ \text{cal} \approx 21\,000\ \text{J}\)
  5. intermediate

    200 g of water at 80 °C are mixed with 300 g of water at 20 °C in an insulated container. What is the equilibrium temperature? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(C_1 = 200 \cdot 1 = 200\ \text{cal}/{}^\circ\text{C}\); \(C_2 = 300 \cdot 1 = 300\ \text{cal}/{}^\circ\text{C}\).
    \(T_{eq} = \dfrac{C_1 T_1 + C_2 T_2}{C_1 + C_2}\)
    \(T_{eq} = \dfrac{200 \cdot 80 + 300 \cdot 20}{500}\) \(= \dfrac{16\,000 + 6000}{500}\) \(= 44\ ^\circ\text{C}\)
    \(T_{eq} = 44\ ^\circ\text{C}\)
  6. intermediate

    A 100 g aluminium block (\(c = 0.22\ \text{cal/g}\cdot{}^\circ\text{C}\)) and 100 g of water (\(c = 1\ \text{cal/g}\cdot{}^\circ\text{C}\)) each absorb 440 cal. By how much does the temperature of each change? Which changes more?

    Show solution
    \(\Delta T = \dfrac{Q}{m \cdot c}\)
    Aluminium: \(C = 100 \cdot 0.22 = 22\ \text{cal}/{}^\circ\text{C}\) \(\Rightarrow \Delta T = \dfrac{440}{22}\) \(= 20\ ^\circ\text{C}\)
    Water: \(C = 100 \cdot 1 = 100\ \text{cal}/{}^\circ\text{C}\) \(\Rightarrow \Delta T = \dfrac{440}{100}\) \(= 4.4\ ^\circ\text{C}\)
    The smaller the heat capacity, the larger the temperature change.
    Aluminium: \(+20\ ^\circ\text{C}\); water: \(+4.4\ ^\circ\text{C}\). The aluminium changes more.
  7. intermediate

    Why does a clinical thermometer need to stay in the armpit for a few minutes before you read it? Why is it made small?

    Show solution
    The thermometer only shows the body's temperature once it reaches thermal equilibrium with it.
    Until then, heat flows from the body to the thermometer and the reading is still rising.
    Being small (low heat capacity), it needs little heat to warm up and hardly cools the skin.
    You must wait for thermal equilibrium; the small size makes this quick and does not change the temperature being measured.
  8. intermediate

    A 200 g piece of iron at 300 °C is dropped into 500 g of water at 20 °C, in an insulated container. What is the equilibrium temperature? Use \(c(\text{iron}) = 0.11\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) (there is no boiling).

    Show solution
    \(C_{\text{iron}} = 200 \cdot 0.11 = 22\ \text{cal}/{}^\circ\text{C}\); \(C_{\text{water}} = 500 \cdot 1 = 500\ \text{cal}/{}^\circ\text{C}\).
    \(T_{eq} = \dfrac{22 \cdot 300 + 500 \cdot 20}{22 + 500}\)
    \(T_{eq} = \dfrac{6600 + 10\,000}{522}\) \(= \dfrac{16\,600}{522} \approx 31.8\ ^\circ\text{C}\)
    The water has a much larger heat capacity, so its temperature barely moves.
    \(T_{eq} \approx 31.8\ ^\circ\text{C}\)
  9. challenge

    A calorimeter with heat capacity \(50\ \text{cal}/{}^\circ\text{C}\) contains 250 g of water at 20 °C. A 100 g piece of metal at 120 °C is placed inside, and equilibrium is reached at 25 °C. What is the specific heat of the metal? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\); the system is insulated.

    Show solution
    In an insulated system, the heat released by the metal equals the heat absorbed by the water and the calorimeter together.
    Absorbed: \((250 \cdot 1 + 50) \cdot (25 - 20) = 300 \cdot 5 = 1500\ \text{cal}\)
    Released: \(100 \cdot c \cdot (120 - 25) = 9500 \cdot c\)
    \(9500 \cdot c = 1500\) \(\Rightarrow c = \dfrac{1500}{9500} \approx 0.158\ \text{cal/g}\cdot{}^\circ\text{C}\)
    \(c \approx 0.16\ \text{cal/g}\cdot{}^\circ\text{C}\)
  10. challenge

    For a baby's bath, you have 400 g of water at 15 °C and want to reach 40 °C by mixing in water at 90 °C. How much hot water should you use? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and ignore losses.

    Show solution
    Heat released by the hot water = heat absorbed by the cold water.
    \(m \cdot 1 \cdot (90 - 40) = 400 \cdot 1 \cdot (40 - 15)\)
    \(50 \cdot m = 10\,000\) \(\Rightarrow m = 200\ \text{g}\)
    Check: \(T_{eq} = \dfrac{200 \cdot 90 + 400 \cdot 15}{600}\) \(= \dfrac{24\,000}{600}\) \(= 40\ ^\circ\text{C}\).
    \(m = 200\ \text{g}\) of water at 90 °C
STEP 4

Mechanical equivalent of heat: Joule's experiment

Until the early 19th century, many people thought heat was an invisible fluid called ‘caloric’. Around 1845, James Joule showed that heat is energy, with an experiment in which a falling weight turns paddles inside water, in an insulated container, and the water warms up.

If no heat escapes, the energy the weight loses, \(E = M\,g\,h\), shows up entirely as heat in the water, \(Q = m\,c\,\Delta T\). Here \(c\) is the specific heat of water, and 1 calorie warms 1 gram of water by 1 °C (we will study the \(c\) of other materials in steps 6 and 7). Measuring both, Joule always found the same ratio, with each calorie worth about 4.19 joules, and we call this number the mechanical equivalent of heat.

\(E = M\,g\,h\)\(Q = m\,c\,\Delta T\)\(1\ \text{cal} = 4.186\ \text{J}\)\(E = 4.186\ \tfrac{\text{J}}{\text{cal}} \cdot Q\)The weight goes down slowly, at almost constant speed, so nearly all the potential energy goes into the paddles rather than into the speed of the weight.

Let's discuss

  • Drop the weight once and note how many joules it lost and how many calories the water gained. What do you get when you divide one by the other?
  • Why does the water warm up so little per drop, and what do you suppose Joule had to do to measure the difference?
  • Double the mass of water. What happens to ΔT, and what happens to the J/cal ratio?
  • When you rub your hands, brake a bicycle or hammer a nail, where does the energy of motion turn into heat?
Drops
Mechanical energy E
Heat in the water Q
ΔT of the water
E / Q
Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    In the 18th century, many scientists thought heat was an invisible fluid, ‘caloric’, which flowed from the hot body to the cold one. What did Joule's experiment show that overturned this idea?

    Show solution
    In the experiment, a weight falls and turns paddles inside water in an insulated container. No hotter body touches the water, and yet it warms up.
    If heat were a substance, it could not appear out of nowhere just by stirring the water. What happens is that the potential energy of the weight becomes thermal energy of the water.
    Conclusion: heat is a form of energy. Mechanical work and heat convert into each other at a fixed ratio (\(1\ \text{cal} \approx 4.186\ \text{J}\)), and total energy is conserved.
  2. basic

    A 20 kg weight goes down 2 m in a Joule apparatus. Use \(g = 10\ \text{m/s}^2\) and \(1\ \text{cal} \approx 4.2\ \text{J}\). How much energy does it release, in joules and in calories?

    Show solution
    The potential energy lost by the weight is
    \(E = M\,g\,h = 20 \cdot 10 \cdot 2 = 400\ \text{J}\)
    and, converted to calories,
    \(Q = \dfrac{400}{4.2} \approx 95.2\ \text{cal}\)
    \(E = 400\ \text{J} \approx 95.2\ \text{cal}\)
  3. basic

    Using \(1\ \text{cal} \approx 4.2\ \text{J}\), convert 840 J into calories. Then, using the more precise value \(1\ \text{cal} = 4.186\ \text{J}\), convert 1000 cal into joules.

    Show solution
    To go from joules to calories, we divide
    \(\dfrac{840}{4.2} = 200\ \text{cal}\)
    and to go from calories to joules, we multiply
    \(1000 \cdot 4.186 = 4186\ \text{J}\)
    \(840\ \text{J} = 200\ \text{cal}\) and \(1000\ \text{cal} = 4186\ \text{J}\)
  4. basic

    Explain, without calculations, why your hands get warm when you rub them together and why a nail gets hot after being hit by a hammer several times.

    Show solution
    In both cases mechanical work is being done, by friction between the hands and by the hammer striking and deforming the nail.
    The energy of motion (kinetic) does not disappear; it goes into the jiggling of the molecules of the skin or the metal, that is, it becomes thermal energy.
    Mechanical energy turns into heat, and that is why the temperature rises. This is conservation of energy, the same idea as in Joule's experiment.
  5. intermediate

    In a Joule apparatus, a 10 kg weight falls 2 m, and this is repeated 20 times. The paddles stir 500 g of water in an insulated container. Use \(g = 10\ \text{m/s}^2\), \(c_{\text{water}} = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(1\ \text{cal} \approx 4.2\ \text{J}\). How much does the water warm up?

    Show solution
    One drop gives
    \(E_1 = M\,g\,h = 10 \cdot 10 \cdot 2 = 200\ \text{J}\)
    and 20 drops give
    \(E = 20 \cdot 200 = 4000\ \text{J}\)
    In calories, this is
    \(Q = \dfrac{4000}{4.2} \approx 952.4\ \text{cal}\)
    All this energy goes into the water, so
    \(Q = m\,c\,\Delta T\) \(\Rightarrow \Delta T = \dfrac{952.4}{500 \cdot 1} \approx 1.9\ ^\circ\text{C}\)
    \(\Delta T \approx 1.9\ ^\circ\text{C}\)
  6. intermediate

    The water of a waterfall drops 84 m. Assume all the potential energy becomes heat in the water itself, at the bottom of the fall. Use \(g = 10\ \text{m/s}^2\), \(c_{\text{water}} = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(1\ \text{cal} \approx 4.2\ \text{J}\). How much does the water warm up?

    Show solution
    Take 1 kg of water (= 1000 g), whose potential energy is
    \(E = m\,g\,h = 1 \cdot 10 \cdot 84 = 840\ \text{J}\)
    In calories, this is
    \(Q = \dfrac{840}{4.2} = 200\ \text{cal}\)
    This 1 kg then warms by
    \(\Delta T = \dfrac{Q}{m\,c} = \dfrac{200}{1000 \cdot 1} = 0.2\ ^\circ\text{C}\)
    Notice that the mass cancels out, so any amount of water warms up by the same amount.
    \(\Delta T = 0.2\ ^\circ\text{C}\)
  7. intermediate

    A 1000 kg car at 20 m/s brakes to a stop. Assume all the kinetic energy becomes heat in the brake discs, which add up to 8 kg of steel (\(c = 0.11\ \text{cal/g}\cdot{}^\circ\text{C}\)). Use \(1\ \text{cal} \approx 4.2\ \text{J}\). How much do the discs warm up?

    Show solution
    The kinetic energy of the car is
    \(E_k = \dfrac{m\,v^2}{2} = \dfrac{1000 \cdot 20^2}{2} = 200\,000\ \text{J}\)
    In calories, this is
    \(Q = \dfrac{200\,000}{4.2} \approx 47\,619\ \text{cal}\)
    With 8 kg = 8000 g of steel, the discs warm by
    \(\Delta T = \dfrac{Q}{m\,c}\) \(= \dfrac{47\,619}{8000 \cdot 0.11} \approx 54.1\ ^\circ\text{C}\)
    \(\Delta T \approx 54\ ^\circ\text{C}\)
  8. intermediate

    A packet of biscuits contains 100 Cal (food calories, \(1\ \text{Cal} = 1\ \text{kcal}\)). Use \(1\ \text{cal} = 4.186\ \text{J}\) and \(g = 10\ \text{m/s}^2\). If all this energy were used to lift a 60 kg person, how high would they go?

    Show solution
    We convert to calories and then to joules.
    \(100\ \text{Cal} = 100\ \text{kcal} = 100\,000\ \text{cal}\)
    \(E = 100\,000 \cdot 4.186 = 418\,600\ \text{J}\)
    Setting this equal to the potential energy, we get
    \(E = m\,g\,h\) \(\Rightarrow h = \dfrac{418\,600}{60 \cdot 10} \approx 697.7\ \text{m}\)
    (In practice the body uses only part of it; the rest becomes heat.)
    \(h \approx 698\ \text{m}\)
  9. challenge

    A 10 g lead bullet travelling at 300 m/s gets stuck in a block of wood and stops. Half of the kinetic energy becomes heat in the bullet (\(c_{\text{lead}} = 0.03\ \text{cal/g}\cdot{}^\circ\text{C}\)). Use \(1\ \text{cal} \approx 4.2\ \text{J}\). By how much does the temperature of the bullet rise?

    Show solution
    With 10 g = 0.010 kg, the kinetic energy is
    \(E_k = \dfrac{m\,v^2}{2} = \dfrac{0.010 \cdot 300^2}{2} = 450\ \text{J}\)
    Half of it stays in the bullet,
    \(E_{\text{bullet}} = \dfrac{450}{2} = 225\ \text{J}\)
    In calories, this is
    \(Q = \dfrac{225}{4.2} \approx 53.57\ \text{cal}\)
    and the bullet warms by
    \(\Delta T = \dfrac{Q}{m\,c}\) \(= \dfrac{53.57}{10 \cdot 0.03} \approx 178.6\ ^\circ\text{C}\)
    \(\Delta T \approx 179\ ^\circ\text{C}\)
  10. challenge

    In a Joule apparatus, a 15 kg weight falls 1.5 m each time and stirs 1 kg of water. Use \(g = 9.8\ \text{m/s}^2\) (more precise), \(c_{\text{water}} = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(1\ \text{cal} = 4.186\ \text{J}\). What is the minimum number of drops needed to warm the water by 1 °C?

    Show solution
    The heat needed to warm 1000 g of water by 1 °C is
    \(Q = m\,c\,\Delta T = 1000 \cdot 1 \cdot 1 = 1000\ \text{cal}\)
    In joules:
    \(Q = 1000 \cdot 4.186 = 4186\ \text{J}\)
    Each drop gives
    \(E_1 = M\,g\,h = 15 \cdot 9.8 \cdot 1.5 = 220.5\ \text{J}\)
    so the number of drops is
    \(n = \dfrac{4186}{220.5} \approx 18.98\)
    Since \(n\) must be a whole number, we round up.
    19 drops are needed.
STEP 5

Thermal expansion: heating pushes the particles apart

With more jiggling, the particles take up more space, and the body expands. Each material expands by a different amount, which we measure with the coefficient of linear expansion α.

In one dimension (a bar, a rail, a wire), the change in length is proportional to the initial length and to the change in temperature. For area and volume, the coefficients become 2α and 3α, as long as the expansion stays small.

\(\Delta L = L_0\,\alpha\,\Delta T\)\(\Delta A = A_0\,2\alpha\,\Delta T\)\(\Delta V = V_0\,3\alpha\,\Delta T\)α of steel ≈ 1.2·10⁻⁵ °C⁻¹, so a 1 m bar heated by 100 °C grows only 1.2 mm. The drawing exaggerates this 50 times.

Let's discuss

  • Heat the plate. Does the hole get bigger or smaller? Our first intuition tends to say smaller, so check it carefully and try to explain what you see.
  • Which way does the bimetallic strip bend when it heats up, and which way when it cools down?
  • Why do rails and bridges have expansion joints with gaps?
  • ‘Invar’ barely expands at all. What is a material like that useful for?
  • Water behaves strangely, since between 0 °C and 4 °C it contracts when warmed. That is why lakes freeze only at the surface.
α of the material
ΔL of the bar
Area: increase
Volume: increase
Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A steel rail is 20 m long at 10 °C. By how much does its length increase when the temperature reaches 40 °C? Use \(\alpha (\text{steel}) = 1.2 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\).

    Show solution
    \(\Delta L = L_0\,\alpha \,\Delta T\)
    \(\Delta T = 40 - 10 = 30\ ^\circ\text{C}\)
    \(\Delta L = 20 \cdot 1.2 \cdot 10^{-5} \cdot 30 = 7.2 \cdot 10^{-3}\ \text{m}\)
    \(\Delta L = 0.0072\ \text{m} = 7.2\ \text{mm}\)
  2. basic

    An aluminium plate has an area of 2.0 m² and is heated by 50 °C. By how much does the area increase? Use \(\alpha (\text{aluminium}) = 2.4 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\) and \(\beta = 2\alpha\).

    Show solution
    \(\beta = 2\alpha\) \(= 2 \cdot 2.4 \cdot 10^{-5}\) \(= 4.8 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\)
    \(\Delta A = A_0\,\beta \,\Delta T\) \(= 2.0 \cdot 4.8 \cdot 10^{-5} \cdot 50\)
    \(\Delta A = 4.8 \cdot 10^{-3}\ \text{m}^2\)
    \(\Delta A = 0.0048\ \text{m}^2 = 48\ \text{cm}^2\)
  3. basic

    A metal plate has a circular hole in the middle. When the plate is heated, does the hole get bigger, smaller or stay the same? Explain.

    Show solution
    When heated, all the dimensions of the plate grow in the same proportion, like a photographic enlargement.
    The hole behaves as if it were made of the same material as the plate, so its edge also moves away from the centre.
    The hole gets bigger (it expands as if it were filled with the metal itself).
  4. basic

    A copper bar is 100 cm long at 0 °C. What is its length at 100 °C? Use \(\alpha (\text{copper}) = 1.7 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\).

    Show solution
    \(\Delta L = L_0\,\alpha \,\Delta T\) \(= 100 \cdot 1.7 \cdot 10^{-5} \cdot 100\)
    \(\Delta L = 0.17\ \text{cm}\)
    \(L = L_0 + \Delta L = 100 + 0.17\)
    \(L = 100.17\ \text{cm}\)
  5. intermediate

    A solid steel block has a volume of 1000 cm³ at 20 °C and is heated to 220 °C. By how much does its volume increase? Use \(\alpha (\text{steel}) = 1.2 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\) and \(\gamma = 3\alpha\).

    Show solution
    \(\gamma = 3\alpha\) \(= 3 \cdot 1.2 \cdot 10^{-5}\) \(= 3.6 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\)
    \(\Delta T = 220 - 20 = 200\ ^\circ\text{C}\)
    \(\Delta V = V_0\,\gamma \,\Delta T\) \(= 1000 \cdot 3.6 \cdot 10^{-5} \cdot 200\)
    \(\Delta V = 7.2\ \text{cm}^3\)
  6. intermediate

    A bimetallic strip is made of steel (\(\alpha = 1.2 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\)) bonded to aluminium (\(\alpha = 2.4 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\)). When heated, which way does it bend? How is this used in a thermostat?

    Show solution
    Aluminium expands twice as much as steel for the same \(\Delta T\), so the aluminium side becomes longer than the steel side.
    To fit one side that is longer than the other, the strip bends, with the aluminium on the outside (convex) and the steel on the inside.
    In a thermostat (clothes iron, blinking lights), the strip touches a contact and closes the circuit; when it heats up it bends, moves away and cuts the current; when it cools it bends back and reconnects.
    It bends towards the steel side (the metal that expands less), opening/closing the circuit automatically.
  7. intermediate

    A steel bridge is 500 m long. Over the year, the temperature varies from −5 °C to 35 °C. What is the maximum change in length that the expansion joints must absorb? Use \(\alpha (\text{steel}) = 1.2 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\).

    Show solution
    \(\Delta T = 35 - (-5) = 40\ ^\circ\text{C}\)
    \(\Delta L = L_0\,\alpha \,\Delta T\) \(= 500 \cdot 1.2 \cdot 10^{-5} \cdot 40\)
    \(\Delta L = 0.24\ \text{m}\)
    Without the joints, the bridge would buckle or crack as it expanded.
    \(\Delta L = 0.24\ \text{m} = 24\ \text{cm}\)
  8. intermediate

    A 2.0 m bar grows 2.4 mm when heated by 50 °C. Calculate the coefficient of linear expansion and say which metal it is made of: steel (\(1.2 \cdot 10^{-5}\)), aluminium (\(2.4 \cdot 10^{-5}\)) or copper (\(1.7 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\)).

    Show solution
    \(\Delta L = 2.4\ \text{mm} = 2.4 \cdot 10^{-3}\ \text{m}\)
    \(\alpha = \dfrac{\Delta L}{L_0\,\Delta T}\) \(= \dfrac{2.4 \cdot 10^{-3}}{2.0 \cdot 50}\)
    \(\alpha = \dfrac{2.4 \cdot 10^{-3}}{100}\) \(= 2.4 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\)
    \(\alpha = 2.4 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\) → aluminium
  9. challenge

    A 60 L steel tank is filled to the brim with petrol on a cold day and then warms up by 20 °C in the sun. How much petrol overflows? Use \(\gamma (\text{petrol}) = 1.1 \cdot 10^{-3}\ ^\circ\text{C}^{-1}\) and \(\alpha (\text{steel}) = 1.2 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\).

    Show solution
    Petrol: \(\Delta V_{\text{petrol}} = 60 \cdot 1.1 \cdot 10^{-3} \cdot 20 = 1.32\ \text{L}\)
    Tank (the container also expands): \(\gamma (\text{steel}) = 3 \cdot 1.2 \cdot 10^{-5}\) \(= 3.6 \cdot 10^{-5}\ ^\circ\text{C}^{-1}\)
    \(\Delta V_{\text{tank}} = 60 \cdot 3.6 \cdot 10^{-5} \cdot 20 = 0.0432\ \text{L}\)
    Apparent expansion \(= \Delta V_{\text{petrol}} - \Delta V_{\text{tank}}\) \(= 1.32 - 0.0432\) \(= 1.2768\ \text{L}\)
    About \(1.28\ \text{L}\) of petrol overflows
  10. challenge

    In a harsh winter, a lake freezes only at the surface and the fish survive at the bottom, at about 4 °C. Explain this using the anomalous behaviour of water.

    Show solution
    Almost every substance contracts when it cools, but water is different, expanding from 4 °C down to 0 °C and reaching its maximum density at 4 °C.
    As the lake cools, surface water at 4 °C, being denser, sinks; water between 4 °C and 0 °C, being less dense, stays on top.
    The top layer freezes; the ice (less dense) floats and acts as a thermal insulator.
    Water at 4 °C (the densest) stays at the bottom and the ice insulates the surface, so the lake does not freeze all the way through.
STEP 6

Heat capacity and specific heat

The heat capacity C tells you how many calories a body must absorb to rise by 1 °C, and it depends on size, so a pot full of water has a much bigger C than a cup.

The specific heat c tells you how many calories 1 gram of the material needs to rise by 1 °C. Since c is a property of the material, like a fingerprint, it stays the same whatever the size of the body. An aluminium pot and an aluminium spoon have the same c and very different values of C.

Think of a bucket. If heat capacity is the width of the bucket and heat is the water we pour in, then the temperature change is the height the water reaches. Pour the same amount into two buckets, and the level rises a little in the wide one and a lot in the narrow one.

\(C = \dfrac{Q}{\Delta T}\)\(c = \dfrac{Q}{m\,\Delta T}\)\(C = m\,c\)C in cal/°C (property of the body) · c in cal/g·°C (property of the material) · 1 cal/g·°C ≈ 4200 J/(kg·K)
Specific heat of some materials
Materialc (cal/g·°C)
Water1.00
Cooking oil0.50
Ice0.50
Aluminium0.22
Glass0.20
Iron0.11
Copper0.093
Lead0.031

Let's discuss

  • In ‘Same mass, different materials’, which one heats up more with the same heat, and what does that suggest about c?
  • In ‘Pot and spoon’, is c the same? Is C? Which one would you expect to heat up first on the hob?
  • In ‘Same heat capacity’, 55 g of water and 500 g of iron rise together. Why?
  • Why do seaside cities have steadier temperatures than desert cities?
C of A
C of B
c of A
c of B
ΔT of A
ΔT of B
Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A large pot and a small spoon are made of the same aluminium. Do they have the same specific heat? And the same heat capacity? Explain.

    Show solution
    Specific heat (\(c\)) is a property of the MATERIAL, and since both are aluminium, both have \(c = 0.22\ \text{cal/g}\cdot{}^\circ\text{C}\).
    Heat capacity (\(C = m\,c\)) is a property of the BODY and depends on mass.
    The pot has a much bigger mass \(\Rightarrow C_{\text{pot}} > C_{\text{spoon}}\).
    Same specific heat; different heat capacities (the pot has a bigger \(C\)).
  2. basic

    A body absorbs 600 cal and its temperature rises by 20 °C. What is the heat capacity of this body?

    Show solution
    \(C = \dfrac{Q}{\Delta T}\)
    \(C = \dfrac{600\ \text{cal}}{20\ ^\circ\text{C}}\)
    \(C = 30\ \text{cal}/{}^\circ\text{C}\)
  3. basic

    Calculate the heat capacity of a 500 g iron block. Use \(c_{\text{iron}} = 0.11\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(C = m\,c\)
    \(C = 500\ \text{g} \cdot 0.11\ \text{cal/g}\cdot{}^\circ\text{C}\)
    \(C = 55\ \text{cal}/{}^\circ\text{C}\)
  4. basic

    Coastal cities have little temperature difference between day and night, while in the desert the variation is huge. Explain using specific heat (\(c_{\text{water}} = 1\ \text{cal/g}\cdot{}^\circ\text{C}\), much higher than that of sand).

    Show solution
    Water has a high specific heat, so it must absorb a lot of heat to warm up and gives off a lot of heat to cool down.
    On the coast, the sea absorbs heat during the day without warming much and releases this heat at night → moderate temperatures.
    In the desert there is almost no water; the sand, with a low \(c\), heats up a lot during the day and cools quickly at night.
    The high \(c\) of water ‘holds’ the coastal temperature steady; in the desert, without water, the variation is large.
  5. intermediate

    A 200 g sample of an unknown metal absorbs 1320 cal and warms by 30 °C. Calculate the specific heat and identify the metal in the table: aluminium 0.22; glass 0.20; iron 0.11; copper 0.093; lead 0.031 (cal/g·°C).

    Show solution
    \(c = \dfrac{Q}{m\,\Delta T}\)
    \(c = \dfrac{1320}{200 \cdot 30}\)
    \(c = \dfrac{1320}{6000} = 0.22\ \text{cal/g}\cdot{}^\circ\text{C}\)
    In the table, \(0.22\ \text{cal/g}\cdot{}^\circ\text{C}\) corresponds to aluminium.
    \(c = 0.22\ \text{cal/g}\cdot{}^\circ\text{C}\) → aluminium
  6. intermediate

    A copper block and a lead block, both 100 g, each absorb 300 cal. How much does each warm up, and which warms more? Use \(c_{\text{copper}} = 0.093\) and \(c_{\text{lead}} = 0.031\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(\Delta T = \dfrac{Q}{m\,c}\)
    Copper: \(\Delta T = \dfrac{300}{100 \cdot 0.093}\) \(= \dfrac{300}{9.3} \approx 32.3\ ^\circ\text{C}\)
    Lead: \(\Delta T = \dfrac{300}{100 \cdot 0.031}\) \(= \dfrac{300}{3.1} \approx 96.8\ ^\circ\text{C}\)
    With the same \(Q\) and the same mass, the one with the smaller \(c\) warms more (lead warms 3 times as much).
    Copper \(\approx 32.3\ ^\circ\text{C}\); lead \(\approx 96.8\ ^\circ\text{C}\) → lead warms more
  7. intermediate

    Why do we use water in a hot-water bottle and in a car radiator, and not oil (\(c_{\text{oil}} = 0.5\ \text{cal/g}\cdot{}^\circ\text{C}\))? Take \(c_{\text{water}} = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    Water has twice the specific heat of oil.
    Hot-water bottle: to cool by 1 °C, each gram of water gives off twice as much heat → the bottle stays warm for longer.
    Radiator: each gram of water absorbs a lot of heat from the engine while warming only a little → it cools the engine efficiently.
    Because its \(c\) is high, water stores and carries a lot of heat with little change in temperature.
  8. intermediate

    Convert to \(\text{J/(kg}\cdot\text{K)}\) the specific heat of water (\(1\ \text{cal/g}\cdot{}^\circ\text{C}\)) and of aluminium (\(0.22\ \text{cal/g}\cdot{}^\circ\text{C}\)). Use \(1\ \text{cal} \approx 4.2\ \text{J}\), \(1\ \text{g} = 10^{-3}\ \text{kg}\) and a change of \(1\ ^\circ\text{C}\) = a change of \(1\ \text{K}\).

    Show solution
    \(1\ \text{cal/g}\cdot{}^\circ\text{C} = \dfrac{4.2\ \text{J}}{10^{-3}\ \text{kg} \cdot 1\ \text{K}}\) \(= 4200\ \text{J/(kg}\cdot\text{K)}\)
    Water: \(1 \cdot 4200 = 4200\ \text{J/(kg}\cdot\text{K)}\)
    Aluminium: \(0.22 \cdot 4200 = 924\ \text{J/(kg}\cdot\text{K)}\)
    Water \(\approx 4200\ \text{J/(kg}\cdot\text{K)}\); aluminium \(\approx 924\ \text{J/(kg}\cdot\text{K)}\)
  9. challenge

    A 400 g aluminium pot contains 1000 g of water, all at 20 °C. What is the heat capacity of the whole set, and how much heat is needed to bring it to 70 °C? Use \(c_{\text{aluminium}} = 0.22\) and \(c_{\text{water}} = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(C_{\text{pot}} = 400 \cdot 0.22 = 88\ \text{cal}/{}^\circ\text{C}\)
    \(C_{\text{water}} = 1000 \cdot 1 = 1000\ \text{cal}/{}^\circ\text{C}\)
    \(C_{\text{total}} = 88 + 1000 = 1088\ \text{cal}/{}^\circ\text{C}\)
    \(\Delta T = 70 - 20 = 50\ ^\circ\text{C}\)
    \(Q = C_{\text{total}} \cdot \Delta T = 1088 \cdot 50 = 54\,400\ \text{cal}\)
    Notice that about 92% of this heat goes into the water.
    \(C = 1088\ \text{cal}/{}^\circ\text{C}\); \(Q = 54\,400\ \text{cal}\) (\(54.4\ \text{kcal}\))
  10. challenge

    What mass of water has the same heat capacity as a 2 kg iron block? If each one absorbs 4400 cal, how much will each warm up? Use \(c_{\text{iron}} = 0.11\) and \(c_{\text{water}} = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(C_{\text{iron}} = m\,c\) \(= 2000\ \text{g} \cdot 0.11\) \(= 220\ \text{cal}/{}^\circ\text{C}\)
    For the water, \(m \cdot 1 = 220\) \(\Rightarrow m = 220\ \text{g}\)
    With the same heat capacity, the same heat gives the same \(\Delta T\), namely
    \(\Delta T = \dfrac{Q}{C} = \dfrac{4400}{220} = 20\ ^\circ\text{C}\)
    220 g of water; both warm up by 20 °C
STEP 7

Sensible heat: which one heats up fastest?

When heat changes the temperature of a body, we call it sensible heat, and how much of it is needed depends on the mass, the temperature change and the material.

The specific heat c tells you how many calories are needed to heat 1 g of the material by 1 °C. Water has a huge c, so it needs a lot of heat to warm up, and that is also why it takes a long time to cool down.

\(Q = m\,c\,\Delta T\)\(C = m\,c\)\(P = \dfrac{Q}{\Delta t}\)c in cal/g·°C: water 1 · oil 0.5 · aluminium 0.22 · iron 0.11 · copper 0.093 · 1 cal ≈ 4.2 J

Let's discuss

  • They all receive the same heat per second. Which one heats up fastest, and how might that relate to c?
  • On the T × t graph, what does the slope of each line mean?
  • Double the mass. What happens to the slopes?
  • At the beach, during the day, the sand heats up much more than the sea. How does this create the sea breeze?
Time
Heat given to each
T of the water
C = m·c of the water
Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    How much heat is needed to heat 200 g of water from 20 °C to 70 °C? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(Q = m\,c\,\Delta T\)
    \(\Delta T = 70 - 20 = 50\ ^\circ\text{C}\)
    \(Q = 200 \cdot 1 \cdot 50\)
    \(Q = 10\,000\ \text{cal}\)
  2. basic

    What is the heat capacity of a 500 g aluminium block? Use \(c(\text{aluminium}) = 0.22\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(C = m\,c = 500 \cdot 0.22\)
    In other words, the block needs 110 cal to rise by 1 °C.
    \(C = 110\ \text{cal}/{}^\circ\text{C}\)
  3. basic

    A 300 g iron block absorbs 1650 cal. By how much does its temperature rise? Use \(c(\text{iron}) = 0.11\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(\Delta T = \dfrac{Q}{m\,c}\)
    \(m\,c = 300 \cdot 0.11 = 33\ \text{cal}/{}^\circ\text{C}\)
    \(\Delta T = \dfrac{1650}{33}\)
    \(\Delta T = 50\ ^\circ\text{C}\)
  4. basic

    At the beach, at noon, the sand gets hot while the sea water stays cool, even under the same sun. Why? And what is the effect on the wind?

    Show solution
    Water has a specific heat (\(1\ \text{cal/g}\cdot{}^\circ\text{C}\)) much higher than sand, so it needs much more heat to rise each degree.
    Under the same sun, the sand heats up faster; the air above it warms, becomes less dense and rises.
    The cooler air above the sea moves in to take its place, and this is the sea breeze (from sea to land) during the day.
    Water's specific heat is high → sand heats up faster → the breeze blows from sea to land during the day.
  5. intermediate

    A 500 W immersion heater heats 1000 g of water from 20 °C to 100 °C. Assuming all the energy goes into the water, how long does it take? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(1\ \text{cal} \approx 4.2\ \text{J}\).

    Show solution
    \(Q = m\,c\,\Delta T = 1000 \cdot 1 \cdot 80 = 80\,000\ \text{cal}\)
    In joules, this is \(80\,000 \cdot 4.2 = 336\,000\ \text{J}\)
    \(\Delta t = \dfrac{Q}{P} = \dfrac{336\,000}{500} = 672\ \text{s}\)
    \(\Delta t = 672\ \text{s} \approx 11.2\ \text{min}\)
  6. intermediate

    A 400 g aluminium pot contains 1000 g of water at 25 °C. A hob supplies 200 cal/s to the whole set. How long does it take to reach 75 °C? Use \(c(\text{aluminium}) = 0.22\) and \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(C(\text{pot}) = 400 \cdot 0.22 = 88\ \text{cal}/{}^\circ\text{C}\)
    \(C(\text{water}) = 1000 \cdot 1 = 1000\ \text{cal}/{}^\circ\text{C}\)
    \(C(\text{total}) = 1088\ \text{cal}/{}^\circ\text{C}\)
    \(Q = C\,\Delta T = 1088 \cdot 50 = 54\,400\ \text{cal}\)
    \(\Delta t = \dfrac{Q}{P} = \dfrac{54\,400}{200}\)
    \(\Delta t = 272\ \text{s} \approx 4.5\ \text{min}\)
  7. intermediate

    On the \(T \times Q\) graph of a 400 g body, the temperature goes from 10 °C to 60 °C when it absorbs 2000 cal (a straight line). Calculate the heat capacity, the specific heat and the slope of the line.

    Show solution
    \(C = \dfrac{Q}{\Delta T} = \dfrac{2000}{50} = 40\ \text{cal}/{}^\circ\text{C}\)
    \(c = \dfrac{C}{m} = \dfrac{40}{400} = 0.1\ \text{cal/g}\cdot{}^\circ\text{C}\)
    \(\text{Slope} = \dfrac{\Delta T}{Q} = \dfrac{50}{2000} = \dfrac{1}{C}\)
    \(C = 40\ \text{cal}/{}^\circ\text{C}\); \(c = 0.1\ \text{cal/g}\cdot{}^\circ\text{C}\); \(\text{slope} = 0.025\ {}^\circ\text{C}/\text{cal}\)
  8. intermediate

    100 g of water and 100 g of copper absorb 186 cal each. By how much does the temperature of each rise? How many times more does the copper warm up? Use \(c(\text{water}) = 1\) and \(c(\text{copper}) = 0.093\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    Copper: \(\Delta T = \dfrac{186}{100 \cdot 0.093}\) \(= \dfrac{186}{9.3}\) \(= 20\ ^\circ\text{C}\)
    Water: \(\Delta T = \dfrac{186}{100 \cdot 1} = 1.86\ ^\circ\text{C}\)
    The ratio is \(\dfrac{20}{1.86} = \dfrac{1}{0.093} \approx 10.75\)
    Copper: \(20\ ^\circ\text{C}\); water: \(1.86\ ^\circ\text{C}\); the copper warms up \(\approx 10.8\) times more
  9. challenge

    An 840 W microwave heats 250 g of water at 20 °C for 60 s. Assuming all the energy goes into the water, what is the final temperature? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(1\ \text{cal} = 4.2\ \text{J}\).

    Show solution
    \(E = P\,\Delta t = 840 \cdot 60 = 50\,400\ \text{J}\)
    \(Q = \dfrac{50\,400}{4.2} = 12\,000\ \text{cal}\)
    \(\Delta T = \dfrac{Q}{m\,c} = \dfrac{12\,000}{250} = 48\ ^\circ\text{C}\)
    \(T = 20 + 48\)
    \(T = 68\ ^\circ\text{C}\)
  10. challenge

    A 1000 W immersion heater runs for 7 min and only 80% of the energy goes into the water, which goes from 20 °C to 100 °C. What is the mass of water? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(1\ \text{cal} = 4.2\ \text{J}\).

    Show solution
    \(E = P\,\Delta t = 1000 \cdot 420 = 420\,000\ \text{J}\)
    The useful part is \(0.8 \cdot 420\,000 = 336\,000\ \text{J}\)
    \(Q = \dfrac{336\,000}{4.2} = 80\,000\ \text{cal}\)
    \(m = \dfrac{Q}{c\,\Delta T} = \dfrac{80\,000}{1 \cdot 80}\)
    \(m = 1000\ \text{g} = 1\ \text{kg}\)
STEP 8

Phase change: heat that does not change the temperature

Heat ice at −20 °C over a steady flame, and the temperature rises to 0 °C and then… stops, something that may surprise you. The ice keeps melting and the thermometer does not move until the last little piece disappears. Then it rises again up to 100 °C and stops once more, this time while the water boils.

Particles attract each other. In ice, each one is held to its neighbours by firm bonds and can only vibrate in place. Liquid water has weak bonds that form and break all the time, letting the particles slide past one another, while steam has practically no bonds at all.

On the plateaus, the heat is used to break bonds, overcoming the attraction between particles. This energy is stored as interaction (potential) energy and does not increase the jiggling, and since temperature measures the jiggling, the temperature stays put. We call this heat latent heat. Boiling costs more than melting because the particles have to be pulled completely apart.

\(Q = m\,L\)\(L_{\text{fusion}} = 80\ \text{cal/g}\)\(L_{\text{vaporisation}} = 540\ \text{cal/g}\)c: ice 0.5 · water 1 · steam 0.5 cal/g·°C. Boiling 1 g of water takes almost 7 times the heat needed to melt it.

Let's discuss

  • Which part of the graph is the longest, and what does that tell you about boiling water?
  • During melting, watch the bonds. What happens to them while the temperature stays the same?
  • Compare the two ‘where did the heat go’ bars. In which stretches does each one grow?
  • Why does evaporating sweat cool the body?
  • In a pressure cooker, water boils above 100 °C. Why does that cook food faster?
Temperature
What is happening
Total heat supplied
In this stage
Ice bonds intact
Exercises for step 8 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    How much heat is needed to melt 50 g of ice at 0 °C? Use \(L_{\text{fusion}} = 80\ \text{cal/g}\).

    Show solution
    \(Q = m\,L = 50 \cdot 80\)
    The temperature stays at 0 °C throughout the melting.
    \(Q = 4000\ \text{cal}\)
  2. basic

    How much heat is needed to vaporise 20 g of water that is already at 100 °C? Use \(L_{\text{vaporisation}} = 540\ \text{cal/g}\).

    Show solution
    \(Q = m\,L = 20 \cdot 540\)
    \(Q = 10\,800\ \text{cal}\)
  3. basic

    Name each transition: (a) ice melting; (b) dry ice turning into gas; (c) the bathroom mirror fogging up; (d) water turning into ice in the freezer; (e) clothes drying on the line.

    Show solution
    (a) solid → liquid: melting (fusion)
    (b) solid → gas directly: sublimation
    (c) vapour → liquid: condensation
    (d) liquid → solid: freezing (solidification)
    (e) liquid → vapour: vaporisation (evaporation)
    Melting, sublimation, condensation, freezing, vaporisation
  4. basic

    Why do beans cook faster in a pressure cooker? And why, in a very high city (such as La Paz), does water boil below 100 °C?

    Show solution
    The boiling point depends on the pressure on the liquid, and more pressure means it boils at a higher temperature.
    In a pressure cooker the pressure inside is higher, water boils at about 120 °C and the food cooks faster.
    At altitude the air pressure is lower, so water boils below 100 °C (about 88 °C in La Paz) and cooking takes longer.
    More pressure → boiling above 100 °C; less pressure (altitude) → boiling below 100 °C.
  5. intermediate

    How much heat is needed to turn 100 g of ice at −20 °C into water at 20 °C? Use \(c(\text{ice}) = 0.5\ \text{cal/g}\cdot{}^\circ\text{C}\), \(L_{\text{fusion}} = 80\ \text{cal/g}\) and \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    1) Ice \(-20\) \(\Rightarrow 0\ ^\circ\text{C}\): \(Q_1 = 100 \cdot 0.5 \cdot 20 = 1000\ \text{cal}\)
    2) Melting at 0 °C: \(Q_2 = 100 \cdot 80 = 8000\ \text{cal}\)
    3) Water \(0\) \(\Rightarrow 20\ ^\circ\text{C}\): \(Q_3 = 100 \cdot 1 \cdot 20 = 2000\ \text{cal}\)
    \(Q = 1000 + 8000 + 2000\)
    \(Q = 11\,000\ \text{cal}\)
  6. intermediate

    How much heat is needed to turn 200 g of water at 30 °C into steam at 100 °C? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(L_{\text{vaporisation}} = 540\ \text{cal/g}\).

    Show solution
    1) Heat \(30\) \(\Rightarrow 100\ ^\circ\text{C}\): \(Q_1 = 200 \cdot 1 \cdot 70 = 14\,000\ \text{cal}\)
    2) Vaporise at 100 °C: \(Q_2 = 200 \cdot 540 = 108\,000\ \text{cal}\)
    \(Q = 14\,000 + 108\,000\)
    Notice that vaporising uses almost 8 times more than heating.
    \(Q = 122\,000\ \text{cal}\)
  7. intermediate

    During a race, an athlete evaporates 100 g of sweat. How much heat does this remove from the body, in calories and in joules? Use \(L_{\text{vaporisation}} = 540\ \text{cal/g}\) and \(1\ \text{cal} \approx 4.2\ \text{J}\).

    Show solution
    To evaporate, the water must absorb heat, and it takes this heat from the skin.
    \(Q = m\,L = 100 \cdot 540 = 54\,000\ \text{cal}\)
    In joules, this is \(54\,000 \cdot 4.2 = 226\,800\ \text{J}\)
    \(Q = 54\,000\ \text{cal} \approx 226\,800\ \text{J}\) removed from the body (that is why sweat cools you)
  8. intermediate

    A source supplies 500 cal/s to 100 g of ice at 0 °C. For how long does the thermometer stay stuck at 0 °C? Use \(L_{\text{fusion}} = 80\ \text{cal/g}\).

    Show solution
    While the ice is melting, all the heat goes into the phase change and the temperature does not change.
    \(Q = m\,L = 100 \cdot 80 = 8000\ \text{cal}\)
    \(\Delta t = \dfrac{Q}{P} = \dfrac{8000}{500}\)
    \(\Delta t = 16\ \text{s}\) at 0 °C
  9. challenge

    50 g of ice at −20 °C turn into steam at 120 °C. Calculate the heat for each stretch, the total and the percentage used in vaporisation. Use \(c(\text{ice}) = 0.5\); \(c(\text{water}) = 1\); \(c(\text{steam}) = 0.5\ \text{cal/g}\cdot{}^\circ\text{C}\); \(L_f = 80\ \text{cal/g}\); \(L_v = 540\ \text{cal/g}\).

    Show solution
    \(Q_1\) ice \(-20\) \(\Rightarrow 0\ ^\circ\text{C}\): \(50 \cdot 0.5 \cdot 20 = 500\ \text{cal}\)
    \(Q_2\) melting: \(50 \cdot 80 = 4000\ \text{cal}\)
    \(Q_3\) water \(0\) \(\Rightarrow 100\ ^\circ\text{C}\): \(50 \cdot 1 \cdot 100 = 5000\ \text{cal}\)
    \(Q_4\) vaporisation: \(50 \cdot 540 = 27\,000\ \text{cal}\)
    \(Q_5\) steam \(100\) \(\Rightarrow 120\ ^\circ\text{C}\): \(50 \cdot 0.5 \cdot 20 = 500\ \text{cal}\)
    \(\text{Total} = 37\,000\ \text{cal}\)
    Vaporisation: \(\dfrac{27\,000}{37\,000} \approx 0.730\)
    \(Q = 37\,000\ \text{cal}\); vaporisation is \(\approx 73\%\) of the total
  10. challenge

    A constant-power source takes 4 min to melt 200 g of ice at 0 °C. After that, how long does it take for the water to go from 0 °C to 100 °C? And to vaporise all this water? Use \(L_f = 80\ \text{cal/g}\), \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(L_v = 540\ \text{cal/g}\).

    Show solution
    Melting: \(Q = 200 \cdot 80 = 16\,000\ \text{cal}\) in 240 s → \(P = \dfrac{16\,000}{240} \approx 66.7\ \text{cal/s}\)
    Heating: \(Q = 200 \cdot 1 \cdot 100 = 20\,000\ \text{cal}\) \(\Rightarrow \Delta t = \dfrac{20\,000}{66.7}\) \(= 300\ \text{s}\) \(= 5\ \text{min}\)
    Vaporising: \(Q = 200 \cdot 540 = 108\,000\ \text{cal}\) \(\Rightarrow \Delta t = \dfrac{108\,000}{66.7}\) \(= 1620\ \text{s}\) \(= 27\ \text{min}\)
    (Shortcut: \(\dfrac{540}{80} = 6.75\) and \(6.75 \cdot 4\ \text{min} = 27\ \text{min}\))
    Heating: 5 min; vaporising: 27 min
STEP 9

Calorimeter: the heat one loses, the other gains

A calorimeter is an insulated container, like a vacuum flask. If we assume no heat gets in or out, everything the hot body releases is absorbed by the cold body.

When we add up the heats with their signs (positive for the one that absorbs, negative for the one that releases), the total is zero. With ice, remember the latent heat as well, since it may happen that not all the ice melts, and then the mixture stays at 0 °C.

\(\sum Q = 0\)\(Q_{\text{released}} = Q_{\text{absorbed}}\)Each term is m·c·ΔT (sensible heat) or m·L (phase change).

Let's discuss

  • Mix hot and cold water in equal masses. Is the final temperature the average?
  • When we drop hot metal into water, why does the water barely warm up?
  • With ice, increase the mass of ice until some ice is left at the end. What is the final temperature then?
  • The bars for heat released and heat absorbed are always the same size. Why?
Final temperature
Heat released
Heat absorbed
Ice at the end
Exercises for step 9 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    200 g of water at 80 °C are mixed with 300 g of water at 20 °C in an ideal calorimeter. What is the equilibrium temperature? Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\).

    Show solution
    \(\sum Q = 0\) \(\Rightarrow 200 \cdot 1 \cdot (T - 80) + 300 \cdot 1 \cdot (T - 20) = 0\)
    \(200T - 16\,000 + 300T - 6000 = 0\)
    \(500T = 22\,000\) \(\Rightarrow T = 44\ ^\circ\text{C}\)
    \(T \approx 44\ ^\circ\text{C}\)
  2. basic

    In an ideal calorimeter, a hot body and a cold body are put in contact. Explain what it means to write \(\sum Q = 0\).

    Show solution
    An ideal calorimeter does not exchange heat with the surroundings, so all the energy that leaves the hot body goes into the cold body.
    Heat released is negative (\(Q < 0\)) and heat absorbed is positive (\(Q > 0\)).
    Adding everything, we get \(Q_{\text{released}} + Q_{\text{absorbed}} = 0\), that is, \(|Q_{\text{released}}| = Q_{\text{absorbed}}\).
    The exchange stops when both reach the same temperature (thermal equilibrium).
    \(\sum Q = 0\): the heat one body loses is exactly the heat the other gains.
  3. basic

    How much heat is needed to melt 50 g of ice that is already at 0 °C? Use \(L_{\text{fusion}} = 80\ \text{cal/g}\).

    Show solution
    During melting the temperature does not change, and \(Q = m\,L\)
    \(Q = 50 \cdot 80 = 4000\ \text{cal}\)
    \(Q = 4000\ \text{cal}\)
  4. basic

    Equal masses of water, one at 10 °C and the other at 50 °C, are mixed in an ideal calorimeter. Without a long calculation, what is the final temperature? Why?

    Show solution
    Since the masses and the specific heat are equal, each degree the hot water loses corresponds to one degree the cold water gains.
    So the final temperature ends up exactly in the middle: \(\dfrac{10 + 50}{2} = 30\ ^\circ\text{C}\).
    Checking with \(m = 100\ \text{g}\), we get \(100 \cdot (T - 10) + 100 \cdot (T - 50) = 0\) \(\Rightarrow T = 30\ ^\circ\text{C}\)
    \(T = 30\ ^\circ\text{C}\) (the average, because the \(m\,c\) values are equal)
  5. intermediate

    A 200 g aluminium block at 100 °C is placed in 400 g of water at 20 °C in an ideal calorimeter. Use \(c(\text{aluminium}) = 0.22\) and \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\). What is the equilibrium temperature?

    Show solution
    Aluminium: \(m\,c = 200 \cdot 0.22 = 44\ \text{cal}/{}^\circ\text{C}\). Water: \(m\,c = 400 \cdot 1 = 400\ \text{cal}/{}^\circ\text{C}\)
    \(\sum Q = 0\) \(\Rightarrow 44 \cdot (T - 100) + 400 \cdot (T - 20) = 0\)
    \(44T - 4400 + 400T - 8000 = 0\)
    \(444T = 12\,400\) \(\Rightarrow T = 27.93\ ^\circ\text{C}\)
    \(T \approx 27.9\ ^\circ\text{C}\)
  6. intermediate

    A 300 g piece of copper at 150 °C is dropped into 500 g of water at 25 °C, in an ideal calorimeter. Use \(c(\text{copper}) = 0.093\) and \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\). What is the final temperature?

    Show solution
    Copper: \(m\,c = 300 \cdot 0.093 = 27.9\ \text{cal}/{}^\circ\text{C}\). Water: \(500\ \text{cal}/{}^\circ\text{C}\)
    \(\sum Q = 0\) \(\Rightarrow 27.9 \cdot (T - 150) + 500 \cdot (T - 25) = 0\)
    \(27.9T - 4185 + 500T - 12\,500 = 0\)
    \(527.9T = 16\,685\) \(\Rightarrow T = 31.61\ ^\circ\text{C}\)
    \(T \approx 31.6\ ^\circ\text{C}\)
  7. intermediate

    50 g of ice at 0 °C are placed in 400 g of water at 30 °C, in an ideal calorimeter. Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(L_{\text{fusion}} = 80\ \text{cal/g}\). Does all the ice melt? What is the final temperature?

    Show solution
    To melt all the ice we need \(Q = 50 \cdot 80 = 4000\ \text{cal}\)
    The water, cooling to 0 °C, could release at most \(400 \cdot 1 \cdot 30 = 12\,000\ \text{cal} \gt 4000\ \text{cal}\) → all the ice melts.
    \(\sum Q = 0\) \(\Rightarrow 50 \cdot 80 + 50 \cdot 1 \cdot (T - 0) + 400 \cdot 1 \cdot (T - 30) = 0\)
    \(4000 + 50T + 400T - 12\,000 = 0\) \(\Rightarrow 450T = 8000\)
    \(T = 17.78\ ^\circ\text{C}\)
    All the ice melts; \(T \approx 17.8\ ^\circ\text{C}\)
  8. intermediate

    A calorimeter with heat capacity 20 cal/°C contains 200 g of water, both at 20 °C. 100 g of water at 80 °C are added. Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\). What is the equilibrium temperature?

    Show solution
    The calorimeter also warms up, and its heat is \(Q = C\,\Delta T\), with \(C = 20\ \text{cal}/{}^\circ\text{C}\) (equivalent to 20 g of water).
    \(\sum Q = 0\) \(\Rightarrow 20 \cdot (T - 20) + 200 \cdot 1 \cdot (T - 20) + 100 \cdot 1 \cdot (T - 80) = 0\)
    \(220 \cdot (T - 20) + 100 \cdot (T - 80) = 0\)
    \(220T - 4400 + 100T - 8000 = 0\) \(\Rightarrow 320T = 12\,400\)
    \(T = 38.75\ ^\circ\text{C}\)
  9. challenge

    200 g of ice at 0 °C are placed in 300 g of water at 40 °C, in an ideal calorimeter. Use \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(L_{\text{fusion}} = 80\ \text{cal/g}\). What is the final temperature and how much ice is left?

    Show solution
    The most heat the water can release going down to 0 °C is \(300 \cdot 1 \cdot 40 = 12\,000\ \text{cal}\)
    while melting all the ice takes \(200 \cdot 80 = 16\,000\ \text{cal}\)
    Since \(12\,000 < 16\,000\), the ice cannot all melt, and the system stops at 0 °C with ice and water.
    The mass that melts is \(m = \dfrac{12\,000}{80} = 150\ \text{g}\)
    and the ice left is \(200 - 150 = 50\ \text{g}\)
    \(T = 0\ ^\circ\text{C}\), 50 g of ice are left (and 450 g of water)
  10. challenge

    A calorimeter with heat capacity 50 cal/°C contains 300 g of water at 50 °C. 100 g of ice at −10 °C are placed in it. Use \(c(\text{ice}) = 0.5\), \(c(\text{water}) = 1\ \text{cal/g}\cdot{}^\circ\text{C}\) and \(L_{\text{fusion}} = 80\ \text{cal/g}\). What is the final temperature?

    Show solution
    Ice from −10 °C to 0 °C: \(Q_1 = 100 \cdot 0.5 \cdot 10 = 500\ \text{cal}\)
    Melting: \(Q_2 = 100 \cdot 80 = 8000\ \text{cal}\) → total to become water at 0 °C: \(8500\ \text{cal}\)
    Water + calorimeter going down to 0 °C release at most \((300 + 50) \cdot 50 = 17\,500\ \text{cal} \gt 8500\) → all the ice melts.
    \(\sum Q = 0\) \(\Rightarrow 8500 + 100 \cdot 1 \cdot (T - 0) + 300 \cdot 1 \cdot (T - 50) + 50 \cdot (T - 50) = 0\)
    \(8500 + 100T + 350T - 17\,500 = 0\) \(\Rightarrow 450T = 9000\)
    \(T = 20\ ^\circ\text{C}\)
STEP 10

How heat travels: conduction, convection and radiation

In conduction, the jiggling passes from particle to particle, without the particles leaving their places. Metals conduct very well, while wood, air and polystyrene conduct poorly and act as insulators.

Convection happens in liquids and gases, where the heated part becomes less dense and rises while the cold part sinks, forming currents that carry heat along with the matter.

In radiation, heat travels as an electromagnetic wave (infrared) and does not need a medium, which is how the Sun warms the Earth through the vacuum of space. Dark surfaces absorb more of it.

Conduction: \(\Phi = \dfrac{k\,A\,\Delta T}{L}\)Φ = heat flow (W) · k = thermal conductivity (W/m·K): copper 400 · aluminium 237 · steel 50 · glass 1 · wood 0.15

Let's discuss

  • Conduction: switch copper for wood. Why do pans have wooden handles?
  • Convection: turn off the flame and watch what happens to the currents. Why is the freezer at the top of the fridge?
  • Radiation: why do dark clothes get hotter in the sun?
  • A vacuum flask blocks all three. How does it manage that? (Vacuum between the walls, mirrored wall, lid.)
  • Does a blanket ‘warm you up’, or does it just stop your body heat from escaping?
Exercises for step 10 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    Why are pans usually made of metal, but their handles of wood or plastic?

    Show solution
    In conduction, thermal jiggling passes from particle to particle through the material.
    Metals are good conductors (they have free electrons that carry energy quickly), so the base heats up fast and passes the heat to the food.
    Wood and plastic are poor conductors (insulators), and hardly any heat reaches your hand.
    Metal conducts heat well for cooking; the insulating handle protects your hand.
  2. basic

    Why is an air conditioner installed high on the wall, and a heater near the floor?

    Show solution
    Cold air is denser and sinks; hot air is less dense and rises.
    Air conditioner up high: the cold air sinks, the warm room air rises to the unit and gets cooled.
    Heater down low: the heated air rises and the cold air sinks to be heated.
    Convection currents form that mix the air of the whole room.
    Both take advantage of convection: cold sinks, hot rises.
  3. basic

    Between the Sun and the Earth there is practically a vacuum. How does the Sun's heat reach us?

    Show solution
    Conduction and convection need matter (particles) to carry energy, and in a vacuum there is no medium.
    Thermal radiation is made of electromagnetic waves (visible light, infrared...), which travel without needing a medium.
    By radiation: electromagnetic waves cross the vacuum.
  4. basic

    People often say a wool jumper ‘warms you up’. Is that correct from a physics point of view?

    Show solution
    The jumper does not produce heat; the body does (metabolism).
    Wool is a poor conductor and traps still air between its fibres, which is also an insulator.
    So the jumper reduces the heat lost from the body to the surroundings.
    No: the jumper does not heat you, it insulates and reduces the loss of the heat your body produces.
  5. intermediate

    A glass window has an area of 2 m² and a thickness of 4 mm. The temperature is 25 °C inside and 15 °C outside. Using Fourier's law \(\Phi = \dfrac{k\,A\,\Delta T}{L}\), with \(k(\text{glass}) = 0.8\ \text{W}/(\text{m} \cdot \text{K})\), calculate the heat flow.

    Show solution
    Data: \(k = 0.8\ \text{W}/(\text{m} \cdot \text{K})\), \(A = 2\ \text{m}^2\), \(\Delta T = 25 - 15 = 10\ ^\circ\text{C}\) (\(= 10\ \text{K}\)), \(L = 4\ \text{mm} = 0.004\ \text{m}\)
    \(\Phi = \dfrac{0.8 \cdot 2 \cdot 10}{0.004}\)
    \(\Phi = \dfrac{16}{0.004} = 4000\ \text{W}\)
    \(\Phi = 4000\ \text{W}\) (from the hot side to the cold side)
  6. intermediate

    Explain why, at the beach, the breeze blows from the sea to the land during the day and from the land to the sea at night.

    Show solution
    During the day: the sand (land) heats up faster than the water (water has a high specific heat). The air over the land gets warmer and rises, and the cooler air over the sea moves in to take its place → sea breeze (sea → land).
    At night: the land cools faster than the water. Now the air over the sea is the warmer one and rises; the air from the land moves out to sea → land breeze (land → sea).
    These are convection currents caused by land and water heating and cooling differently.
  7. intermediate

    Why, in traditional fridges, is the freezer at the top? And why shouldn't you cover the glass or wire shelves?

    Show solution
    The air in contact with the freezer cools, becomes denser and sinks, cooling the food.
    The warmer air from below rises to the freezer and gets cooled, and a convection current forms.
    Open shelves let the air circulate; covering them blocks convection and the fridge cools less well.
    A freezer at the top keeps convection going: cold air sinks, warm air rises.
  8. intermediate

    On a cold morning, you touch a metal door handle and a wooden door, both at the same temperature as the room. The metal feels colder. Why?

    Show solution
    Both are at the same temperature (in equilibrium with the room).
    The feeling of cold depends on how quickly heat leaves your hand.
    Metal is a good conductor and pulls heat from your hand quickly, whereas wood is a poor conductor and pulls heat slowly.
    The metal is not colder; it conducts heat away from your hand faster.
  9. challenge

    A vacuum flask has double glass walls with a vacuum between them, mirrored surfaces and a tightly closed lid. Explain how each detail hinders one mode of heat transfer.

    Show solution
    Vacuum between the walls: with no particles, there is no conduction or convection through the wall.
    Mirrored surfaces: they reflect electromagnetic waves (infrared), reducing radiation.
    Closed lid (made of insulating material): it stops convection of hot air out of the opening and reduces conduction through the lid.
    The vacuum flask fights all three: conduction and convection (vacuum and lid) and radiation (mirroring).
  10. challenge

    A brick wall has an area of 10 m², a thickness of 20 cm and \(k = 0.6\ \text{W}/(\text{m} \cdot \text{K})\); it is 25 °C inside and 10 °C outside. Use \(\Phi = \dfrac{k\,A\,\Delta T}{L}\). Calculate the heat flow, the energy that crosses the wall in 1 hour and the new flow if the thickness were doubled.

    Show solution
    \(\Delta T = 25 - 10 = 15\ ^\circ\text{C}\); \(L = 0.2\ \text{m}\)
    \(\Phi = \dfrac{0.6 \cdot 10 \cdot 15}{0.2} = \dfrac{90}{0.2} = 450\ \text{W}\)
    In \(1\ \text{h} = 3600\ \text{s}\): \(E = \Phi \,\Delta t\) \(= 450 \cdot 3600\) \(= 1\,620\,000\ \text{J} \approx 1.62 \cdot 10^{6}\ \text{J}\)
    Doubling \(L\) to 0.4 m: \(\Phi ^{\prime} = \dfrac{90}{0.4} = 225\ \text{W}\) (the flow is halved, because \(\Phi\) is inversely proportional to \(L\))
    \(\Phi = 450\ \text{W}\); \(E \approx 1.62 \cdot 10^{6}\ \text{J}\) in 1 h; with double the thickness, \(\Phi = 225\ \text{W}\)
STEP 11

Ideal gases: pressure, volume and temperature

In a gas, the particles fly around freely and hit the walls. Each hit pushes on the wall, and all these pushes added together make up the pressure, so more hits, or harder hits, mean more pressure.

With less volume, the particles hit more often and the pressure goes up. A higher temperature makes them hit faster and harder, which also raises the pressure (or the gas pushes the piston and its volume grows).

The particles also hit one another and swap energy, which is why their speeds follow the Maxwell distribution (histogram below the cylinder). Heating the gas stretches the curve towards higher speeds.

\(P\,V = n\,R\,T\)\(\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}\)R = 0.082 atm·L/(mol·K) · T always in kelvin · isothermal: P·V constant · isobaric: V/T constant · isochoric: P/T constant

Let's discuss

  • Isothermal: halve the volume. What happens to the pressure, and what curve appears on the P × V graph?
  • Isochoric: heat the gas. Why is a tyre pumped up in the morning fuller in the afternoon?
  • Isobaric: cool the gas. Why does a balloon shrink in the freezer?
  • In each of the three cases, does P·V/T change? Check the number in the readings.
Pressure P
Volume V
Temperature T
P·V/T
Exercises for step 11 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A gas occupies 6 L at a pressure of 2 atm. It is slowly compressed, at constant temperature, to 3 L. What is the new pressure?

    Show solution
    Isothermal process (Boyle's law): \(P_1\,V_1 = P_2\,V_2\)
    \(2 \cdot 6 = P_2 \cdot 3\)
    \(P_2 = \dfrac{12}{3} = 4\ \text{atm}\)
    \(P_2 = 4\ \text{atm}\) (the volume halved, the pressure doubled)
  2. basic

    A gas occupies 4 L at 27 °C. Keeping the pressure constant, it is heated to 177 °C. What is the new volume?

    Show solution
    Convert to kelvin: \(T_1 = 27 + 273 = 300\ \text{K}\); \(T_2 = 177 + 273 = 450\ \text{K}\)
    Isobaric: \(\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\)
    \(\dfrac{4}{300} = \dfrac{V_2}{450}\) \(\Rightarrow V_2 = \dfrac{4 \cdot 450}{300} = 6\ \text{L}\)
    \(V_2 = 6\ \text{L}\)
  3. basic

    A tyre is pumped up in the morning to a pressure of 2.0 atm at 27 °C. In the afternoon, driving on hot asphalt, the air inside reaches 57 °C. Taking the volume of the tyre as constant, what is the new pressure?

    Show solution
    \(T_1 = 27 + 273 = 300\ \text{K}\); \(T_2 = 57 + 273 = 330\ \text{K}\)
    Isochoric (constant volume): \(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\)
    \(\dfrac{2.0}{300} = \dfrac{P_2}{330}\) \(\Rightarrow P_2 = \dfrac{2.0 \cdot 330}{300} = 2.2\ \text{atm}\)
    That is why tyres should be pumped up when they are cold.
    \(P_2 = 2.2\ \text{atm}\)
  4. basic

    A classmate doubled the temperature of a gas from 20 °C to 40 °C, at constant volume, and said the pressure doubled. Is she right? Why do we always use kelvin?

    Show solution
    The gas laws hold with absolute temperature (kelvin), which starts at absolute zero, where the particles' motion would be at its minimum.
    \(20\ ^\circ\text{C} = 293\ \text{K}\) and \(40\ ^\circ\text{C} = 313\ \text{K}\), so the absolute temperature increased only 1.068 times.
    So \(\dfrac{P_2}{P_1} = \dfrac{313}{293} \approx 1.068\), and the pressure increased by about 6.8% instead of doubling.
    She is not right: in kelvin the temperature did not double, and the pressure rises only \(\approx 6.8\%\).
  5. intermediate

    A 10 L container holds 2 mol of an ideal gas at 27 °C. Use \(P\,V = n\,R\,T\) with \(R = 0.082\ \text{atm} \cdot \text{L}/(\text{mol} \cdot \text{K})\). What is the pressure of the gas?

    Show solution
    \(T = 27 + 273 = 300\ \text{K}\)
    \(P = \dfrac{n\,R\,T}{V} = \dfrac{2 \cdot 0.082 \cdot 300}{10}\)
    \(P = \dfrac{49.2}{10} = 4.92\ \text{atm}\)
    \(P = 4.92\ \text{atm}\)
  6. intermediate

    Show, using \(P\,V = n\,R\,T\) with \(R = 0.082\ \text{atm} \cdot \text{L}/(\text{mol} \cdot \text{K})\), that 1 mol of ideal gas at STP (0 °C and 1 atm) occupies about 22.4 L. Then calculate the volume of 3 mol at STP.

    Show solution
    \(T = 0 + 273 = 273\ \text{K}\); \(P = 1\ \text{atm}\); \(n = 1\ \text{mol}\)
    \(V = \dfrac{n\,R\,T}{P}\) \(= \dfrac{1 \cdot 0.082 \cdot 273}{1}\) \(= 22.386\ \text{L} \approx 22.4\ \text{L}\)
    For 3 mol: \(V = 3 \cdot 22.4 = 67.2\ \text{L}\) (using the formula: \(3 \cdot 22.386 \approx 67.16\ \text{L}\))
    1 mol \(\approx 22.4\ \text{L}\); 3 mol \(\approx 67.2\ \text{L}\)
  7. intermediate

    A full party balloon is put in the freezer and, after a while, it has shrunk. Explain using the particle model.

    Show solution
    In the freezer the air in the balloon cools, and the particles start moving more slowly.
    Being slower, they hit the balloon's wall less hard and less often, and the pressure inside tends to drop.
    The rubber (and the air outside) push in and the balloon shrinks until the pressure inside balances the pressure outside again.
    Since the pressure stays practically equal to the outside pressure, it is almost an isobaric process: \(V/T\) constant → smaller \(T\), smaller \(V\).
    Slower particles hit less often and less hard; the volume shrinks with the temperature.
  8. intermediate

    A syringe has its tip sealed and contains 20 mL of air at 1 atm. The plunger is pushed slowly until the air takes up 5 mL, without changing the temperature. What is the final pressure? Why does it get hard to push?

    Show solution
    Isothermal (Boyle): \(P_1\,V_1 = P_2\,V_2\)
    \(1 \cdot 20 = P_2 \cdot 5\) \(\Rightarrow P_2 = 4\ \text{atm}\)
    With less space, the same particles (same speed, because \(T\) is constant) hit the walls more often, raising the pressure.
    \(P_2 = 4\ \text{atm}\)
  9. challenge

    A fixed mass of gas is at 1 atm, occupies 6 L and has a temperature of 27 °C. It changes to 4 L and 127 °C. Use \(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\). What is the new pressure?

    Show solution
    \(T_1 = 27 + 273 = 300\ \text{K}\); \(T_2 = 127 + 273 = 400\ \text{K}\)
    \(P_2 = \dfrac{P_1\,V_1\,T_2}{T_1\,V_2}\)
    \(P_2 = \dfrac{1 \cdot 6 \cdot 400}{300 \cdot 4} = \dfrac{2400}{1200} = 2\ \text{atm}\)
    \(P_2 = 2\ \text{atm}\)
  10. challenge

    A rigid 8.2 L cylinder contains ideal gas at 6 atm and 27 °C. Use \(R = 0.082\ \text{atm} \cdot \text{L}/(\text{mol} \cdot \text{K})\). (a) How many moles are in the cylinder? (b) What is the pressure if the cylinder is heated to 127 °C?

    Show solution
    (a) \(T_1 = 300\ \text{K}\). \(n = \dfrac{P\,V}{R\,T} = \dfrac{6 \cdot 8.2}{0.082 \cdot 300} = \dfrac{49.2}{24.6} = 2\ \text{mol}\)
    (b) \(T_2 = 127 + 273 = 400\ \text{K}\). \(P = \dfrac{n\,R\,T}{V} = \dfrac{2 \cdot 0.082 \cdot 400}{8.2} = 8\ \text{atm}\)
    Checking with the isochoric law: \(P_2 = \dfrac{6 \cdot 400}{300} = 8\ \text{atm}\ \checkmark\)
    (a) \(n = 2\ \text{mol}\); (b) \(P = 8\ \text{atm}\)
WRAP-UP

Challenges

Scales

An American thermometer reads 104 °F. Does the person have a fever?

Show solution
\(\dfrac{C}{5} = \dfrac{F - 32}{9}\) \(\Rightarrow C = \dfrac{(104 - 32) \cdot 5}{9}\)
\(C = \dfrac{72 \cdot 5}{9} = 40\ ^\circ\text{C}\)
Yes, it is a high fever.
Thermal expansion

A 20 m steel rail goes from 10 °C to 50 °C. By how much does it grow? (\(\alpha = 1.2 \cdot 10^{-5}\ {}^\circ\text{C}^{-1}\))

Show solution
\(\Delta L = L_0\,\alpha\,\Delta T\) \(= 20 \cdot 1.2 \cdot 10^{-5} \cdot 40\)
\(\Delta L = 9.6 \cdot 10^{-3}\ \text{m} \approx 9.6\ \text{mm}\)
That is why rails have gaps between them.
Sensible heat

How much heat is needed to heat 2 L of water (2000 g) from 20 °C to 100 °C? Give it in calories and in joules (1 cal ≈ 4.2 J).

Show solution
\(Q = m\,c\,\Delta T = 2000 \cdot 1 \cdot 80 = 160\,000\ \text{cal}\)
In joules, this is \(160\,000 \cdot 4.2 \approx 672\,000\ \text{J} = 672\ \text{kJ}\)
Phase change

How much heat turns 50 g of ice at 0 °C into water at 20 °C? (\(L_f = 80\ \text{cal/g}\))

Show solution
Melt: \(Q_1 = m\,L = 50 \cdot 80 = 4000\ \text{cal}\)
Heat: \(Q_2 = m\,c\,\Delta T = 50 \cdot 1 \cdot 20 = 1000\ \text{cal}\)
Total \(= 5000\ \text{cal}\)
Calorimeter

200 g of water at 80 °C are mixed with 300 g of water at 20 °C. What is the final temperature?

Show solution
\(200 \cdot 1 \cdot (T - 80) + 300 \cdot 1 \cdot (T - 20) = 0\)
\(500\,T = 16\,000 + 6000 = 22\,000\)
\(T = 44\ ^\circ\text{C}\)
Equilibrium

A 100 g piece of iron at 200 °C is dropped into 500 g of water at 20 °C. What is the final temperature? (\(c_{\text{iron}} = 0.11\ \text{cal/g}\cdot{}^\circ\text{C}\))

Show solution
\(C_{\text{iron}} = 100 \cdot 0.11 = 11\ \text{cal}/{}^\circ\text{C}\) · \(C_{\text{water}} = 500\ \text{cal}/{}^\circ\text{C}\)
\(T = \dfrac{500 \cdot 20 + 11 \cdot 200}{500 + 11} = \dfrac{12\,200}{511}\)
\(T \approx 23.9\ ^\circ\text{C}\), and the water barely changes.
Gases

A gas in a rigid container is at 27 °C and 2.0 atm. It is heated to 87 °C. What is the new pressure?

Show solution
Constant volume: \(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\) (in kelvin!)
\(T_1 = 300\ \text{K}\), \(T_2 = 360\ \text{K}\)
\(P_2 = 2.0 \cdot \dfrac{360}{300} = 2.4\ \text{atm}\)
Think, no math

Why is the air conditioner high on the wall and the heater near the floor?

Show solution
Cold air is denser and sinks; hot air is less dense and rises.
When cold air is released up high and warm air down low, convection spreads the temperature through the whole room.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with an everyday context, a question and five options, each one returning to a different step of the lesson. Further down we list real exam questions on the same topics, which may well be the best practice once these are done.

  1. Ideal gases · Step 11

    Rafael checked his car tyres in the morning, when the air inside was at 27 °C and its absolute pressure was 3.0 atm. After an hour on the motorway on a hot day, the air in the tyres reached 57 °C. Assume the tyre volume does not change, no air escapes and the air behaves as an ideal gas.

    Under these conditions, what is the new absolute pressure of the air in the tyres?

    1. 2.7 atm
    2. 3.0 atm
    3. 3.3 atm
    4. 3.6 atm
    5. 6.3 atm
    Show solution
    Answer: C.
    With the volume and the amount of air fixed, the pressure is proportional to the absolute temperature, so we convert everything to kelvin.
    \(T_1 = 27 + 273 = 300\ \text{K}\) and \(T_2 = 57 + 273 = 330\ \text{K}\)
    \(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\) \(\Rightarrow P_2 = 3.0 \cdot \dfrac{330}{300} = 3.3\ \text{atm}\)
    Using degrees Celsius in the ratio gives 6.3 atm, and inverting the ratio gives 2.7 atm. The 3.6 atm comes from comparing 330 K with 273 K instead of 300 K, a slip that seems small and still changes the answer.
    In a real tyre the rubber also gives a little, and the pressure tends to end up slightly below this value.
  2. Expansion · Step 5

    Lúcia knows that a glass jar with a stuck lid will often open after a stream of hot water over the lid. She has identical jars of ordinary glass with lids made of different materials, all 8.0 cm in diameter. To keep things simple, assume the lid and the neck of the jar both warm up by 50 °C.

    Coefficient of linear expansion
    Materialα (10⁻⁵ °C⁻¹)
    Ordinary glass0.9
    Steel1.2
    Copper1.7
    Aluminium2.4

    Which lid should open most easily, and by how much does its diameter then exceed that of the neck?

    1. The steel one, by 0.012 mm.
    2. The copper one, by 0.032 mm.
    3. The aluminium one, by 0.096 mm.
    4. It makes no difference, because the glass also expands and the gap stays at zero.
    5. The aluminium one, by 0.060 mm.
    Show solution
    Answer: E.
    Every lid in the table expands more than the glass, and the gap, which is the difference between how much the lid grows and how much the neck grows, is largest for the material with the highest α, aluminium.
    \(\Delta D = D_0\,(\alpha_{\text{lid}} - \alpha_{\text{glass}})\,\Delta T\) \(= 8.0 \cdot (2.4 - 0.9) \cdot 10^{-5} \cdot 50\)
    \(\Delta D = 6.0 \cdot 10^{-3}\ \text{cm} = 0.060\ \text{mm}\)
    The 0.096 mm shows up when we forget that the neck expands too. The steel and copper lids leave smaller gaps, of 0.012 mm and 0.032 mm.
    In the kitchen the lid heats up faster than the glass, which probably makes the real gap a little larger than in our model.
  3. Phase change · Step 8

    In a school laboratory, Tiago and Yasmin heated 100 g of ice, initially at −20 °C, in a container over a flame of constant power. Assuming the flame's entire output goes into the sample, they obtained the graph below. The specific heat of ice is 0.5 cal/g·°C.

    −200100 1919 T (°C)t (min)

    Based on the graph, what latent heat of fusion of ice did they measure?

    1. 80 cal/g
    2. 90 cal/g
    3. 100 cal/g
    4. 160 cal/g
    5. 8000 cal/g
    Show solution
    Answer: A.
    The first stretch gives us the power of the flame, because there we know how much heat the ice received.
    \(Q_1 = m\,c\,\Delta T = 100 \cdot 0.5 \cdot 20 = 1000\ \text{cal}\) in 1 min, that is, \(P = 1000\ \text{cal/min}\)
    The plateau at 0 °C runs from minute 1 to minute 9 and lasts 8 min, which gives \(Q_f = 1000 \cdot 8 = 8000\ \text{cal}\).
    \(L = \dfrac{Q_f}{m} = \dfrac{8000}{100} = 80\ \text{cal/g}\)
    The result matches the table value, which indicates that ignoring losses was a reasonable assumption here.
    The 90 shows up when we read the end of the plateau (9 min) as its length, and the 100 when we use the 10 min of the water stretch. The 160 comes from taking the specific heat of ice as 1, and the 8000 from forgetting to divide by the mass.
  4. Calorimeter · Step 9

    Helena made 300 g of tea at 90 °C and wants to drink it at 40 °C without waiting. She pours the tea into an insulated cup and adds ice cubes at 0 °C, perhaps the quickest way to cool it. Ignore the heat exchanged with the cup and the air and treat the tea as water (c = 1 cal/g·°C), with a latent heat of fusion of ice of 80 cal/g.

    What mass of ice leaves the mixture at 40 °C once every ice cube has melted?

    1. 88 g
    2. 125 g
    3. 188 g
    4. 225 g
    5. 375 g
    Show solution
    Answer: B.
    The heat the tea gives off as it cools from 90 °C to 40 °C is what the ice receives to melt and then for the meltwater to warm from 0 °C to 40 °C.
    \(Q_{\text{released}} = 300 \cdot 1 \cdot (90 - 40) = 15\,000\ \text{cal}\)
    \(Q_{\text{absorbed}} = m \cdot 80 + m \cdot 1 \cdot (40 - 0) = 120\,m\)
    \(120\,m = 15\,000\) \(\Rightarrow m = 125\ \text{g}\)
    Forgetting to warm the meltwater gives 188 g, and forgetting the latent heat gives 375 g. Using 90 °C as the change for the tea leads to 225 g, and taking the meltwater all the way to 90 °C leads to 88 g.
  5. Heat transfer · Step 10

    Caio left his car closed in the midday sun for two hours. Outside, the air was at 32 °C, and when he came back he found the air inside the car well above 50 °C. Plant greenhouses use the same principle to keep the inside warmer than the outside.

    Which explanation accounts for this warming?

    1. The glass conducts heat from the outside air inwards much better than the bodywork, and the heat builds up inside.
    2. The sun compresses the air inside the car, and the higher pressure makes the temperature rise.
    3. The glass is an insulator and stops the cold outside from getting into the car.
    4. The glass lets visible light through, the seats absorb it and re-emit it as infrared, radiation the glass lets through poorly, and the closed car stops convection from carrying the hot air away.
    5. The dark seats reflect the sunlight, which gets trapped between the windows and heats the air with each reflection.
    Show solution
    Answer: D.
    The energy comes in by radiation, mostly as visible light, which passes through the glass and is absorbed by the seats and the dashboard.
    These warmed objects emit infrared, which ordinary glass lets out poorly, and the hot air is trapped as well, because the closed car prevents convection with the air outside.
    Option C treats cold as something that ‘gets in’, when heat always flows from hot to cold. A reverses the role of the glass, which is a poor conductor, and B and E describe mechanisms that do not occur there.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where we can look each question up by year, day, booklet colour and number.

  • ENEM 2010, Day 1, blue booklet, question 50. It starts from the gap between the everyday and the scientific use of ‘heat’ and ‘temperature’ and asks for the situation that shows the limit of the everyday idea, which is water boiling without changing temperature.
  • ENEM 2015, Day 1, blue booklet, question 65. A vacuum flask receives cold and hot water, and we need the equilibrium temperature of the mixture before rating the flask by how much the temperature drops in six hours.
  • ENEM 2016, Day 1, blue booklet, question 84. Ice cubes are placed on plastic and aluminium trays of equal mass, and the question asks on which one the ice melts faster and why, which depends on thermal conductivity.
  • ENEM 2019, Day 2, blue booklet, question 102. Two insulated containers of different sizes hold identical ice blocks, and from the mass of ice melted in each we work out the ratio between the thermal conductivities of their walls.
  • ENEM 2023, Day 2, blue booklet, question 132. A tyre warmed by a journey cools to room temperature at constant volume, and the question asks for the pressure–temperature graph that describes this cooling.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Scales\(\frac{C}{5} = \frac{F-32}{9} = \frac{K-273}{5}\)
Thermal expansion\(\Delta L = L_0\alpha\Delta T \quad \beta = 2\alpha \quad \gamma = 3\alpha\)
Sensible heat\(Q = m\,c\,\Delta T\)
Heat capacity\(C = Q/\Delta T = m\,c\)
Specific heat\(c = Q/(m\,\Delta T)\)
Mechanical equivalent\(1\ \text{cal} = 4.186\ \text{J}\)
Latent heat\(Q = m\,L\)
Calorimeter\(\sum Q = 0\)
Conduction\(\Phi = k\,A\,\Delta T/L\)
Ideal gases\(P\,V = n\,R\,T\)
Fixed mass of gas\(P_1V_1/T_1 = P_2V_2/T_2\)
Units1 cal ≈ 4.2 J · T in K for gases