Thermal energy transfers HL Papers 1& 2

B.1 — Thermal Energy Transfers
HL Practice: Papers 1 & 2

20 multiple-choice questions • 15 extended-response questions • 1 lab investigation

How to use this practice bank

These original IB-style questions combine core concepts with demanding calculations, explanations, and experimental evaluation for HL students. Attempt each question before opening its answer. The multiple-choice section follows the Paper 1A format; the extended-response section follows the Paper 2 style. This is a topic practice bank, rather than a complete examination paper.

Marks: multiple choice: 20; extended response: 104; lab investigation: 20. Suggested marking guidance appears inside each expandable answer. Show your working and include units in calculated answers. Use unrounded intermediate values and give final calculated results to a sensible precision, usually three significant figures.

Data and equations

Use the following values unless a question states otherwise: $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ , $c_{\mathrm{ice}}=2100\,\mathrm{J\,kg^{-1}\,K^{-1}}$ , $L_f=3.34\times10^5\,\mathrm{J\,kg^{-1}}$ , $\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ , and $\lambda_{\max}T=2.90\times10^{-3}\,\mathrm{m\,K}$ .

$\displaystyle Q=mc\Delta T,\qquad Q=mL,\qquad P=\frac{Q}{t}$ $\displaystyle P_{\mathrm{conduction}}=\frac{kA\Delta T}{d},\qquad P_{\mathrm{emitted}}=\varepsilon\sigma AT^4$ $\displaystyle L_{\mathrm{star}}=4\pi R^2\sigma T^4,\qquad b=\frac{L_{\mathrm{star}}}{4\pi d^2}$ Temperature differences have equal numerical values in Celsius and kelvin. Use absolute temperature in kelvin for radiation laws. Treat heat capacities and other quoted material properties as constant over the specified ranges.

Paper 1A style — Multiple-choice questions

Choose one answer, A–D, for each question. Each question is worth 1 mark.

Question 1 — Temperature scales [1 mark]

Which kelvin temperature is closest to $25.0\,^{\circ}\mathrm{C}$ ?

$25.0\,\mathrm{K}$
$298\,\mathrm{K}$
$248\,\mathrm{K}$
$273\,\mathrm{K}$

Show answer and explanation

Answer: B. $T=25.0+273.15=298.15\,\mathrm{K}$ , so the closest value is $298\,\mathrm{K}$ .

Question 2 — Temperature differences [1 mark]

A sample’s temperature increases by $18\,^{\circ}\mathrm{C}$ . What is the temperature increase in kelvin?

$255\,\mathrm{K}$
$291\,\mathrm{K}$
$0.066\,\mathrm{K}$
$18\,\mathrm{K}$

Show answer and explanation

Answer: D. The kelvin and Celsius scales have equal-sized intervals. A change of $18\,^{\circ}\mathrm{C}$ is a change of $18\,\mathrm{K}$ ; do not add $273.15$ to a temperature difference.

Question 3 — Internal energy [1 mark]

Which statement defines the internal energy of a system?

The total microscopic random kinetic energy and interaction potential energy of its particles
The average kinetic energy of its particles
The energy transferred into it only by heating
The kinetic energy of its centre of mass

Show answer and explanation

Answer: A. Internal energy is microscopic stored energy. It includes particle kinetic and potential energies; heat is an energy transfer, and centre-of-mass motion is a macroscopic contribution.

Question 4 — Density [1 mark]

A solid has mass $0.540\,\mathrm{kg}$ and volume $2.00\times10^{-4}\,\mathrm{m^3}$ . What is its density?

$270\,\mathrm{kg\,m^{-3}}$
$1.08\times10^{-4}\,\mathrm{kg\,m^{-3}}$
$2.70\times10^3\,\mathrm{kg\,m^{-3}}$
$3.70\times10^{-4}\,\mathrm{kg\,m^{-3}}$

Show answer and explanation

Answer: C. $\displaystyle \rho=\frac{m}{V}=\frac{0.540}{2.00\times10^{-4}}=2.70\times10^3\,\mathrm{kg\,m^{-3}}$

Question 5 — Specific heat capacity [1 mark]

A $0.250\,\mathrm{kg}$ block with specific heat capacity $800\,\mathrm{J\,kg^{-1}\,K^{-1}}$ receives $4000\,\mathrm{J}$ without changing phase or losing energy. What is its temperature rise?

$20.0\,\mathrm{K}$
$5.00\,\mathrm{K}$
$12.5\,\mathrm{K}$
$80.0\,\mathrm{K}$

Show answer and explanation

Answer: A. $\displaystyle \Delta T=\frac{Q}{mc}=\frac{4000}{(0.250)(800)}=20.0\,\mathrm{K}$

Question 6 — Heating-rate comparison [1 mark]

An ideal heater warms a block at a certain rate. It is replaced by a block with twice the mass and twice the specific heat capacity. Heater power is unchanged and losses are negligible. What is the new heating rate divided by the original rate?

$4$
$\frac{1}{2}$
$\frac{1}{4}$
$1$

Show answer and explanation

Answer: C. The heating rate is $P/(mc)$ . The product $mc$ becomes four times larger, so the heating rate becomes one quarter.

Question 7 — Melting and particle energy [1 mark]

A pure substance melts at constant pressure while its temperature stays constant. Which statement describes the supplied energy?

It increases the average random kinetic energy only
It leaves the internal energy unchanged
It decreases the interaction potential energy
It changes the intermolecular structure and increases interaction potential energy

Show answer and explanation

Answer: D. During the idealised melting plateau, energy changes the intermolecular arrangement and increases potential energy. Average random kinetic energy and temperature stay approximately constant, while internal energy increases.

Question 8 — Latent heat [1 mark]

What energy melts $0.100\,\mathrm{kg}$ of ice initially at $0\,^{\circ}\mathrm{C}$ ? Use $L_f=3.34\times10^5\,\mathrm{J\,kg^{-1}}$ .

$3.34\times10^2\,\mathrm{J}$
$3.34\times10^4\,\mathrm{J}$
$1.67\times10^4\,\mathrm{J}$
$3.34\times10^5\,\mathrm{J}$

Show answer and explanation

Answer: B. $\displaystyle Q=mL_f=(0.100)(3.34\times10^5)=3.34\times10^4\,\mathrm{J}$

Question 9 — Heating efficiency [1 mark]

A heater draws $200\,\mathrm{W}$ . Only $60\%$ reaches the sample. How long does it take to transfer $72.0\,\mathrm{kJ}$ to the sample?

$216\,\mathrm{s}$
$360\,\mathrm{s}$
$600\,\mathrm{s}$
$900\,\mathrm{s}$

Show answer and explanation

Answer: C. Useful power is $(0.60)(200)=120\,\mathrm{W}$ . Therefore $t=72000/120=600\,\mathrm{s}$ .

Question 10 — Thermal equilibrium [1 mark]

Equal masses of liquid water at $10\,^{\circ}\mathrm{C}$ and $50\,^{\circ}\mathrm{C}$ are mixed in a container of negligible heat capacity. There is no energy transfer to the surroundings. What is the final temperature?

$30\,^{\circ}\mathrm{C}$
$20\,^{\circ}\mathrm{C}$
$40\,^{\circ}\mathrm{C}$
$60\,^{\circ}\mathrm{C}$

Show answer and explanation

Answer: A. Equal masses of the same substance have equal heat capacities, so the final temperature is the arithmetic mean of the initial temperatures.

Question 11 — Incomplete melting [1 mark]

$0.100\,\mathrm{kg}$ of ice at $0\,^{\circ}\mathrm{C}$ is added to $0.200\,\mathrm{kg}$ of water at $30.0\,^{\circ}\mathrm{C}$ . No energy is exchanged with the container or surroundings. Use $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ and $L_f=3.34\times10^5\,\mathrm{J\,kg^{-1}}$ . Which final state is closest?

All the ice melts and the final temperature is $20\,^{\circ}\mathrm{C}$
$0.025\,\mathrm{kg}$ of ice remains, with the mixture at $0\,^{\circ}\mathrm{C}$
All the original water freezes
The full $0.100\,\mathrm{kg}$ of ice remains unchanged

Show answer and explanation

Answer: B. The water can release $(0.200)(4180)(30.0)=25080\,\mathrm{J}$ before reaching $0\,^{\circ}\mathrm{C}$ . This melts $25080/334000=0.0751\,\mathrm{kg}$ , leaving $0.0249\,\mathrm{kg}$ of ice at $0\,^{\circ}\mathrm{C}$ .

Question 12 — Evaporative cooling [1 mark]

Why can evaporation cool the liquid that remains?

Every molecule loses the same amount of kinetic energy
Molecules escaping are always the slowest molecules
Evaporation must occur only at the boiling point
Escaping molecules tend to have greater energy than the average, reducing the average energy of those remaining

Show answer and explanation

Answer: D. Higher-energy molecules are more likely to escape the liquid surface. Their removal lowers the average random kinetic energy of the remaining molecules, unless energy supplied from elsewhere offsets the cooling.

Question 13 — Conduction in metals [1 mark]

Which mechanism contributes to thermal conduction in a metal?

Bulk circulation of the solid
Energy transfer by mobile electrons and lattice vibrations
Electromagnetic radiation travelling through a vacuum only
Matter moving from the hot end to the cold end as a whole

Show answer and explanation

Answer: B. Mobile electrons and interactions between vibrating particles transfer energy through a metal without bulk movement of the solid.

Question 14 — Natural convection [1 mark]

Why does a heated region of fluid often rise?

Its particles stop colliding
Heating always increases its density
Thermal expansion can lower its density, producing an upward buoyant force
Thermal radiation pushes it upwards

Show answer and explanation

Answer: C. A heated fluid often expands and becomes less dense than the cooler surrounding fluid. Buoyancy then drives circulation in a gravitational field.

Question 15 — Conduction and thickness [1 mark]

For steady conduction through a uniform slab, its thickness is doubled while thermal conductivity, area, and the surface-temperature difference remain unchanged. What happens to the energy-transfer rate?

It halves
It doubles
It increases by a factor of four
It stays unchanged

Show answer and explanation

Answer: A. The rate is $P=kA\Delta T/d$ , so doubling $d$ halves the rate.

Question 16 — Conduction calculation [1 mark]

An insulating slab has $k=0.040\,\mathrm{W\,m^{-1}\,K^{-1}}$ , area $2.0\,\mathrm{m^2}$ , thickness $0.10\,\mathrm{m}$ , and a surface-temperature difference of $20\,\mathrm{K}$ . What is the steady conduction rate?

$0.016\,\mathrm{W}$
$0.16\,\mathrm{W}$
$1.6\,\mathrm{W}$
$16\,\mathrm{W}$

Show answer and explanation

Answer: D. $\displaystyle P=\frac{kA\Delta T}{d}=\frac{(0.040)(2.0)(20)}{0.10}=16\,\mathrm{W}$

Question 17 — Stefan–Boltzmann scaling [1 mark]

Two black bodies have equal emitting areas. One has twice the absolute temperature of the other. What is the ratio of their total emitted powers, hotter divided by cooler?

$1$
$2$
$4$
$16$

Show answer and explanation

Answer: D. For equal areas, $P\propto T^4$ , so the ratio is $2^4=16$ . This compares emitted power, not net power exchanged with surroundings.

Question 18 — Inverse-square spreading [1 mark]

A star radiates isotropically. Its apparent brightness is measured at distance $d$ and then at distance $3d$ . Absorption is negligible. What is the second brightness divided by the first?

$\frac{1}{9}$
$\frac{1}{3}$
$3$
$9$

Show answer and explanation

Answer: A. $b\propto1/d^2$ , so the ratio is $d^2/(3d)^2=1/9$ .

Question 19 — Wien’s law [1 mark]

A star’s wavelength spectrum peaks at $500\,\mathrm{nm}$ . Treating the star as a black body, what is its approximate surface temperature? Use $\lambda_{\max}T=2.90\times10^{-3}\,\mathrm{m\,K}$ .

$580\,\mathrm{K}$
$1.45\times10^3\,\mathrm{K}$
$5.80\times10^3\,\mathrm{K}$
$5.80\times10^6\,\mathrm{K}$

Show answer and explanation

Answer: C. $\displaystyle T=\frac{2.90\times10^{-3}}{500\times10^{-9}}=5.80\times10^3\,\mathrm{K}$

Question 20 — Radiative equilibrium [1 mark]

An object is in radiative equilibrium with its surroundings. Which statement is correct?

It stops emitting thermal radiation
It continues emitting and absorbing radiation, with zero net radiative transfer
It emits radiation but absorbs none
Its temperature must be zero kelvin

Show answer and explanation

Answer: B. At equilibrium, absorbed and emitted radiative powers balance. Both processes continue; their net transfer is zero.

Paper 2 style — Extended-response questions

Answer all parts of each question. Bracketed values indicate the suggested marks. Explain physical mechanisms clearly and show the equation used before substitution.

Extended response 1 — Temperature scales [5 marks]

(a) Convert $25\,^{\circ}\mathrm{C}$ to kelvin. [1 mark]

(b) Convert $310\,\mathrm{K}$ to degrees Celsius. [1 mark]

(c) Explain why absolute temperature is used in relationships proportional to temperature, such as the ideal-gas mean translational kinetic-energy equation. [3 marks]

Show full solution and marking guidance

(a) [1 mark] $\displaystyle T=25+273.15=298.15\,\mathrm{K}\approx298\,\mathrm{K}$

(b) [1 mark] $\displaystyle \theta=310-273.15=36.85\,^{\circ}\mathrm{C}\approx37\,^{\circ}\mathrm{C}$

(c) [3 marks] Kelvin is an absolute temperature scale whose zero is absolute zero. Celsius has an arbitrary offset, so Celsius ratios do not express ratios of absolute temperature. For an ideal gas, mean translational kinetic energy is proportional to kelvin temperature. Award one mark for each linked idea; do not require a claim that all particle motion or all internal energy vanishes at absolute zero.

Extended response 2 — Specific heat capacity [5 marks]

(a) A $0.80\,\mathrm{kg}$ aluminium block is heated from $20\,^{\circ}\mathrm{C}$ to $75\,^{\circ}\mathrm{C}$ . State its temperature increase in kelvin. [1 mark]

(b) Calculate the energy absorbed. Use $c=900\,\mathrm{J\,kg^{-1}\,K^{-1}}$ and assume no phase change. [2 marks]

(c) Explain why a Celsius temperature difference can be used numerically in the heat-capacity calculation. [2 marks]

Show full solution and marking guidance

(a) [1 mark] $\displaystyle \Delta T=75-20=55\,\mathrm{K}$

(b) [2 marks] Use $Q=mc\Delta T$ and substitute: $\displaystyle Q=(0.80)(900)(55)=39600\,\mathrm{J}=3.96\times10^4\,\mathrm{J}$

(c) [2 marks] A kelvin interval and a Celsius interval have the same size. The offset of 273.15 cancels when initial temperature is subtracted from final temperature.

Extended response 3 — Melting and internal energy [5 marks]

(a) Calculate the energy required to melt $0.50\,\mathrm{kg}$ of ice initially at $0\,^{\circ}\mathrm{C}$ . Use $L_f=3.34\times10^5\,\mathrm{J\,kg^{-1}}$ . [2 marks]

(b) Explain why the temperature stays constant during melting at constant pressure even though internal energy increases. [3 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle Q=mL_f=(0.50)(3.34\times10^5)=1.67\times10^5\,\mathrm{J}$

(b) [3 marks] Energy changes the intermolecular arrangement, increasing interaction potential energy. Average random kinetic energy remains approximately constant, so temperature stays constant. Internal energy increases because it includes both microscopic kinetic and potential energy. Melting does not split water molecules into separate atoms.

Extended response 4 — Combined warming and melting [7 marks]

(a) Calculate the total energy required to turn $0.40\,\mathrm{kg}$ of ice at $-10\,^{\circ}\mathrm{C}$ into water at $20\,^{\circ}\mathrm{C}$ . Use $c_i=2100\,\mathrm{J\,kg^{-1}\,K^{-1}}$ , $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ , and $L_f=3.34\times10^5\,\mathrm{J\,kg^{-1}}$ . [5 marks]

(b) Explain why a single use of the water heat-capacity equation over the entire temperature range would be incorrect. [2 marks]

Show full solution and marking guidance

(a) [5 marks] Separate the three stages: $\displaystyle Q_1=(0.40)(2100)(10)=8400\,\mathrm{J}$ $\displaystyle Q_2=(0.40)(334000)=133600\,\mathrm{J}$ $\displaystyle Q_3=(0.40)(4180)(20)=33440\,\mathrm{J}$ $\displaystyle Q_{\mathrm{total}}=8400+133600+33440=175440\,\mathrm{J}$ Answer: $1.75\times10^5\,\mathrm{J}$ to three significant figures. Award credit for identifying all three stages, correct heat-capacity and latent-heat terms, and the total.

(b) [2 marks] Ice and water have different specific heat capacities. Melting also requires latent energy while temperature is constant, so that energy is not included in a single $mc\Delta T$ term.

Extended response 5 — Temperature, internal energy, and compression [6 marks]

(a) Distinguish temperature from internal energy. [2 marks]

(b) Two samples of the same ideal gas are at the same temperature. One contains twice as many particles. Compare their mean translational kinetic energies and total translational kinetic energies. [2 marks]

(c) An ideal gas is compressed in a thermally insulated cylinder and its temperature rises. Explain the energy transfer responsible. [2 marks]

Show full solution and marking guidance

(a) [2 marks] Temperature describes thermal state and, for an ideal gas, determines mean random translational kinetic energy. Internal energy is the total microscopic random kinetic and interaction potential energy of the system.

(b) [2 marks] Mean translational kinetic energy per particle is the same, since the temperatures are equal. Total translational kinetic energy is twice as large for the sample containing twice as many particles.

(c) [2 marks] The moving piston does work on the gas, transferring energy to it. With no heat loss, the gas’s internal energy increases; its mean translational kinetic energy and temperature rise.

Extended response 6 — Kettle power and efficiency [7 marks]

(a) A $2.0\,\mathrm{kW}$ kettle heats $1.5\,\mathrm{kg}$ of water from $20\,^{\circ}\mathrm{C}$ to $100\,^{\circ}\mathrm{C}$ . Calculate the energy absorbed by the water. Use $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . [2 marks]

(b) Calculate the minimum heating time, neglecting energy absorbed by the kettle and energy transferred to the surroundings. [2 marks]

(c) State why the measured heating time is likely to be greater. [1 mark]

(d) Calculate the heating time if $80\%$ of the electrical energy reaches the water. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle Q=(1.5)(4180)(100-20)=501600\,\mathrm{J}\approx5.02\times10^5\,\mathrm{J}$

(b) [2 marks] $\displaystyle t_{\min}=\frac{Q}{P}=\frac{501600}{2000}=250.8\,\mathrm{s}\approx251\,\mathrm{s}$

(c) [1 mark] Some energy warms the kettle or is transferred to the surroundings, reducing the power available to warm the water.

(d) [2 marks] $\displaystyle P_{\mathrm{water}}=(0.80)(2000)=1600\,\mathrm{W}$ $\displaystyle t=\frac{501600}{1600}=313.5\,\mathrm{s}\approx314\,\mathrm{s}$

Extended response 7 — Mixing water and the effect of a container [7 marks]

(a) $0.30\,\mathrm{kg}$ of water at $80\,^{\circ}\mathrm{C}$ is mixed with $0.50\,\mathrm{kg}$ of water at $20\,^{\circ}\mathrm{C}$ . Calculate the final temperature if the container’s heat capacity and external losses are negligible. [3 marks]

(b) Recalculate the final temperature if the container initially at $20\,^{\circ}\mathrm{C}$ has heat capacity $60\,\mathrm{J\,K^{-1}}$ . Assume no external loss and use $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . [3 marks]

(c) Explain the direction of the change in final temperature. [1 mark]

Show full solution and marking guidance

(a) [3 marks] Let $\theta$ be the final temperature in degrees Celsius. $\displaystyle 0.30c_w(80-\theta)=0.50c_w(\theta-20)$ $\displaystyle 34=0.80\theta\quad\Rightarrow\quad\theta=42.5\,^{\circ}\mathrm{C}$

(b) [3 marks] Include energy absorbed by the container: $\displaystyle (0.30)(4180)(80-\theta)=[(0.50)(4180)+60](\theta-20)$ $\displaystyle \theta=\frac{(0.30)(4180)(80)+(0.50)(4180)(20)+(60)(20)}{(0.80)(4180)+60}$ $\displaystyle \theta\approx42.1\,^{\circ}\mathrm{C}$

(c) [1 mark] The initially cooler container absorbs energy, so less is available to raise the cold water’s temperature and the equilibrium temperature is lower.

Extended response 8 — Calorimetry with a hot metal [7 marks]

(a) A $0.150\,\mathrm{kg}$ metal sample at $100\,^{\circ}\mathrm{C}$ is placed in $0.250\,\mathrm{kg}$ of water at $20.0\,^{\circ}\mathrm{C}$ . The calorimeter is initially at $20.0\,^{\circ}\mathrm{C}$ and has heat capacity $30.0\,\mathrm{J\,K^{-1}}$ . The final temperature is $25.0\,^{\circ}\mathrm{C}$ . Write an energy-balance equation for the metal’s specific heat capacity, $c_m$ . [3 marks]

(b) Calculate $c_m$ , assuming negligible external energy transfer. Use $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . [2 marks]

(c) Explain the effect on the calculated specific heat capacity if some energy from the metal actually escapes to the surroundings but this is ignored. [2 marks]

Show full solution and marking guidance

(a) [3 marks] Energy lost by metal equals energy gained by water and calorimeter: $\displaystyle (0.150)c_m(100-25)=[(0.250)(4180)+30.0](25-20)$

(b) [2 marks] $\displaystyle c_m=\frac{[(0.250)(4180)+30.0](5.0)}{(0.150)(75)}\approx478\,\mathrm{J\,kg^{-1}\,K^{-1}}$

(c) [2 marks] The model attributes only the measured water-and-calorimeter energy gain to the metal’s energy loss. If additional energy escapes, the true metal energy loss is larger, so the calculated specific heat capacity is underestimated.

Extended response 9 — Interpreting a heating curve [8 marks]

(a) A $100\,\mathrm{W}$ heater warms $0.200\,\mathrm{kg}$ of a pure substance from solid to liquid. Energy losses and container heat capacity are negligible. Using the data below, calculate the solid’s specific heat capacity. [2 marks]

Heating data for a pure substance
Time / s	Temperature / °C
0	−20
40	−10
80	0
280	0
480	0
680	0
760	10
840	20

(b) Calculate the specific latent heat of fusion. [2 marks]

(c) Calculate the liquid’s specific heat capacity. [2 marks]

(d) Explain why the temperature is constant in the middle interval and why the liquid warms more slowly than the solid. [2 marks]

Show full solution and marking guidance

(a) [2 marks] Before melting, temperature rises by $20\,\mathrm{K}$ in $80\,\mathrm{s}$ . $\displaystyle c_s=\frac{Pt}{m\Delta T}=\frac{(100)(80)}{(0.200)(20)}=2000\,\mathrm{J\,kg^{-1}\,K^{-1}}$

(b) [2 marks] The melting interval is $680-80=600\,\mathrm{s}$ . $\displaystyle L_f=\frac{P\Delta t}{m}=\frac{(100)(600)}{0.200}=3.00\times10^5\,\mathrm{J\,kg^{-1}}$

(c) [2 marks] After melting, temperature rises by $20\,\mathrm{K}$ in $160\,\mathrm{s}$ . $\displaystyle c_l=\frac{(100)(160)}{(0.200)(20)}=4000\,\mathrm{J\,kg^{-1}\,K^{-1}}$

(d) [2 marks] During melting, supplied energy changes the intermolecular structure rather than raising average random kinetic energy. The liquid has twice the solid’s specific heat capacity, so its heating rate at the same mass and power is half as large.

Extended response 10 — Ice added to warm water [8 marks]

(a) $0.100\,\mathrm{kg}$ of ice at $-10.0\,^{\circ}\mathrm{C}$ is placed in $0.500\,\mathrm{kg}$ of water at $40.0\,^{\circ}\mathrm{C}$ . Calculate the energy needed to warm the ice to $0\,^{\circ}\mathrm{C}$ . Use $c_i=2100\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . [2 marks]

(b) Calculate the additional energy needed to melt the ice. Use $L_f=3.34\times10^5\,\mathrm{J\,kg^{-1}}$ . [2 marks]

(c) Determine whether all the ice melts and calculate the final temperature. Neglect the container and external losses. Use $c_w=4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . [3 marks]

(d) Explain why assuming a final temperature above the melting point must be checked against the available energy. [1 mark]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle Q_1=(0.100)(2100)(10.0)=2100\,\mathrm{J}$

(b) [2 marks] $\displaystyle Q_2=(0.100)(334000)=33400\,\mathrm{J}$

(c) [3 marks] The warm water can supply $(0.500)(4180)(40.0)=83600\,\mathrm{J}$ before reaching the melting point. This exceeds the $35500\,\mathrm{J}$ needed to warm and melt all the ice. Then: $\displaystyle (0.500)(4180)(40.0-\theta)=35500+(0.100)(4180)\theta$ $\displaystyle \theta=\frac{83600-35500}{(0.600)(4180)}\approx19.2\,^{\circ}\mathrm{C}$

(d) [1 mark] If the available energy were insufficient to melt all the ice, solid and liquid would remain together at the melting temperature, so the assumed all-liquid final state would be invalid.

Extended response 11 — Conduction through a wall and insulation [7 marks]

(a) A wall has area $12.0\,\mathrm{m^2}$ , thickness $0.200\,\mathrm{m}$ , and thermal conductivity $0.800\,\mathrm{W\,m^{-1}\,K^{-1}}$ . Its inner and outer surfaces are held at $293\,\mathrm{K}$ and $273\,\mathrm{K}$ . Calculate the steady conduction rate. [2 marks]

(b) Calculate the energy transferred through the wall in one hour. [2 marks]

(c) A layer of insulation of thickness $0.0500\,\mathrm{m}$ and conductivity $0.0400\,\mathrm{W\,m^{-1}\,K^{-1}}$ is added. The temperature difference across the complete wall-plus-insulation remains $20.0\,\mathrm{K}$ . Calculate the new rate using $P=A\Delta T/(d_1/k_1+d_2/k_2)$ . [2 marks]

(d) Explain one reason why an insulating material containing trapped air can reduce thermal energy transfer. [1 mark]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle P=\frac{kA\Delta T}{d}=\frac{(0.800)(12.0)(20.0)}{0.200}=960\,\mathrm{W}$

(b) [2 marks] $\displaystyle Q=Pt=(960)(3600)=3.46\times10^6\,\mathrm{J}$

(c) [2 marks] $\displaystyle P=\frac{(12.0)(20.0)}{0.200/0.800+0.0500/0.0400}=160\,\mathrm{W}$ The layers carry the same steady energy-transfer rate, while their temperature drops add.

(d) [1 mark] Air has low thermal conductivity, and trapping it in small spaces limits bulk circulation and therefore convection. Credit either correctly explained mechanism.

Extended response 12 — Thermal-transfer mechanisms and net radiation [8 marks]

(a) Explain why a metal absorber connected to water pipes can transfer solar energy to the water effectively by conduction. [2 marks]

(b) Explain how natural convection can circulate water heated from below. [2 marks]

(c) Explain how energy travels from the Sun to the absorber through space, and why conduction and convection cannot account for transfer across a vacuum. [2 marks]

(d) A small surface has emissivity $0.800$ , area $0.0200\,\mathrm{m^2}$ , and temperature $350\,\mathrm{K}$ . It exchanges radiation with large surroundings at $293\,\mathrm{K}$ . Calculate its net outward radiative power using $P_{\mathrm{net}}=\varepsilon\sigma A(T^4-T_s^4)$ , with $\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ . Assume the same effective emissivity for absorption and emission. [2 marks]

Show full solution and marking guidance

(a) [2 marks] The absorber receives radiation and transfers energy along the metal by mobile electrons and lattice interactions. Good thermal contact and the metal’s relatively high thermal conductivity help transfer energy to the pipes and water.

(b) [2 marks] The heated water expands and can become less dense. Buoyancy causes it to rise while cooler, denser water moves down, creating circulation.

(c) [2 marks] Energy travels as electromagnetic radiation, which can propagate through a vacuum. Conduction and convection require matter: conduction transfers energy through interactions within matter, and convection involves bulk movement of a fluid.

(d) [2 marks] $\displaystyle P_{\mathrm{net}}=(0.800)(5.67\times10^{-8})(0.0200)(350^4-293^4)$ $\displaystyle P_{\mathrm{net}}\approx6.93\,\mathrm{W}$

Extended response 13 — Black-body temperature, luminosity, and brightness [7 marks]

(a) A star is modelled as a spherical black body. Its wavelength spectrum peaks at $500\,\mathrm{nm}$ . Calculate its surface temperature using $\lambda_{\max}T=2.90\times10^{-3}\,\mathrm{m\,K}$ . [2 marks]

(b) Its radius is $7.00\times10^8\,\mathrm{m}$ . Calculate its luminosity using $L=4\pi R^2\sigma T^4$ . [2 marks]

(c) Calculate its apparent brightness at distance $1.50\times10^{11}\,\mathrm{m}$ , assuming isotropic emission and negligible absorption. [2 marks]

(d) State the difference between luminosity and apparent brightness. [1 mark]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle T=\frac{2.90\times10^{-3}}{500\times10^{-9}}=5800\,\mathrm{K}$

(b) [2 marks] $\displaystyle L=4\pi(7.00\times10^8)^2(5.67\times10^{-8})(5800)^4\approx3.95\times10^{26}\,\mathrm{W}$

(c) [2 marks] $\displaystyle b=\frac{L}{4\pi d^2}=\frac{3.95\times10^{26}}{4\pi(1.50\times10^{11})^2}\approx1.40\times10^3\,\mathrm{W\,m^{-2}}$

(d) [1 mark] Luminosity is the total power emitted by the star; apparent brightness is received power per unit area at the observer’s location.

Extended response 14 — Inferring stellar distance and uncertainty [9 marks]

(a) A spherical black-body star has radius $4.00\times10^8\,\mathrm{m}$ and peak wavelength $400\,\mathrm{nm}$ . Calculate its temperature. [2 marks]

(b) Calculate its luminosity. [2 marks]

(c) Its measured apparent brightness is $2.00\times10^{-10}\,\mathrm{W\,m^{-2}}$ . Calculate its distance, neglecting absorption. [2 marks]

(d) The radius, peak wavelength, and brightness have percentage uncertainties of $3\%$ , $2\%$ , and $10\%$ , respectively. Estimate the maximum percentage uncertainty in the derived distance by adding absolute fractional contributions. Treat the constants as exact. [3 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle T=\frac{2.90\times10^{-3}}{400\times10^{-9}}=7250\,\mathrm{K}$

(b) [2 marks] $\displaystyle L=4\pi(4.00\times10^8)^2(5.67\times10^{-8})(7250)^4\approx3.15\times10^{26}\,\mathrm{W}$

(c) [2 marks] $\displaystyle d=\sqrt{\frac{L}{4\pi b}}\approx3.54\times10^{17}\,\mathrm{m}$

(d) [3 marks] Combine the equations: $\displaystyle d=R\sqrt{\frac{\sigma T^4}{b}}\propto R\lambda_{\max}^{-2}b^{-1/2}$ Hence: $\displaystyle \frac{\delta d}{d}\approx\frac{\delta R}{R}+2\frac{\delta\lambda}{\lambda}+\frac{1}{2}\frac{\delta b}{b}$ $\displaystyle \text{Maximum percentage uncertainty}=3+2(2)+\frac{10}{2}=12\%$ This is a linear maximum estimate, not a statistical combination in quadrature.

Extended response 15 — Radiative cooling and changing power [8 marks]

(a) A body has mass $0.500\,\mathrm{kg}$ , specific heat capacity $800\,\mathrm{J\,kg^{-1}\,K^{-1}}$ , surface area $0.0300\,\mathrm{m^2}$ , and emissivity $0.750$ . Initially it is at $400\,\mathrm{K}$ in large surroundings at $300\,\mathrm{K}$ . Calculate its initial net outward radiative power using $P_{\mathrm{net}}=\varepsilon\sigma A(T^4-T_s^4)$ . Neglect other energy transfers. [2 marks]

(b) Calculate the magnitude of its initial temperature-change rate. [2 marks]

(c) Explain why this initial rate cannot be treated as constant throughout cooling to the surroundings’ temperature. [2 marks]

(d) Explain what happens to emission, absorption, and net radiative transfer at thermal equilibrium. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle P_{\mathrm{net}}=(0.750)(5.67\times10^{-8})(0.0300)(400^4-300^4)\approx22.3\,\mathrm{W}$

(b) [2 marks] $\displaystyle \left|\frac{\Delta T}{\Delta t}\right|_{\mathrm{initial}}=\frac{P_{\mathrm{net}}}{mc}=\frac{22.3256}{(0.500)(800)}\approx0.0558\,\mathrm{K\,s^{-1}}$ The body’s temperature-change rate is negative because it cools.

(c) [2 marks] As temperature falls, emitted power decreases while absorption from constant-temperature surroundings remains approximately unchanged. The net outward power and cooling-rate magnitude therefore decrease.

(d) [2 marks] The body still emits and absorbs radiation. At equilibrium, those powers balance and net radiative transfer is zero.

Lab investigation — Heating with energy losses

Investigation 1 — Determine a metal’s specific heat capacity [20 marks]

A student heats the same metal block at two different electrical powers. The block has mass $m=(0.500\pm0.005)\,\mathrm{kg}$ . Heater powers are $P_1=(100\pm1)\,\mathrm{W}$ and $P_2=(160\pm1)\,\mathrm{W}$ . Each run starts at $29.00\,^{\circ}\mathrm{C}$ , and the ambient conditions and insulation are unchanged. Over this short temperature interval, approximate the energy-loss rate as constant and equal in both runs. Initially neglect the heat capacities of the heater and probe and assume the block is at a uniform temperature.

Temperature measurements for the two heating runs
Time / s	Run 1 temperature / °C	Run 2 temperature / °C
0	29.00	29.00
2	29.30	29.54
4	29.60	30.08
6	29.90	30.62
8	30.20	31.16
10	30.50	31.70

(a) Describe a practical method to obtain the data. Identify the independent and dependent variables, important controls, and how heater power is measured. [4 marks]

(b) Determine the temperature–time gradient for each run, including units. [3 marks]

(c) Assume a constant energy-loss rate $P_{\mathrm{loss}}$ , identical in both runs. Starting with $P_i=mc s_i+P_{\mathrm{loss}}$ , where $s_i$ is the temperature–time gradient, derive an expression for $c$ and calculate it. [4 marks]

(d) Calculate the energy-loss rate using either run. [2 marks]

(e) The fitted gradients are $s_1=(0.150\pm0.010)\,\mathrm{K\,s^{-1}}$ and $s_2=(0.270\pm0.010)\,\mathrm{K\,s^{-1}}$ . Estimate the maximum percentage uncertainty in $c$ by adding absolute uncertainty contributions. State $c$ with an absolute uncertainty. [3 marks]

(f) Evaluate two limitations of the method and propose a specific improvement for each. [4 marks]

Show investigation solution and evaluation

(a) [4 marks] Use a metal block with an electric heater and temperature probe in close-fitting holes, a low-voltage supply, voltmeter, ammeter, balance, and timer or data logger. Measure the block mass and record temperature against time at two measured heater powers, $P=VI$ . The independent variable is heater power; the dependent quantity is the local temperature–time gradient. Keep the same block, probe position, insulation, and ambient conditions. Begin each run at the same temperature and compare gradients near the same block temperature. Repeat the runs. Use good thermal contact and allow the block to become approximately uniform in temperature. Handle the heated block carefully and keep within the heater’s rated power. Award credit for apparatus and power measurement, variables, controls, and a repeatable recording method.

(b) [3 marks] Use a best-fit straight line for each nearly linear run, with widely separated points on the line: $\displaystyle s_1=\frac{30.50-29.00}{10.0}=0.150\,\mathrm{K\,s^{-1}}$ $\displaystyle s_2=\frac{31.70-29.00}{10.0}=0.270\,\mathrm{K\,s^{-1}}$ Temperature differences have the same numerical value in kelvin and Celsius.

(c) [4 marks] Subtract the energy-balance equations to eliminate the common loss rate: $\displaystyle P_2-P_1=mc(s_2-s_1)$ $\displaystyle c=\frac{P_2-P_1}{m(s_2-s_1)}=\frac{160-100}{(0.500)(0.270-0.150)}=1.00\times10^3\,\mathrm{J\,kg^{-1}\,K^{-1}}$ Credit the subtraction, rearrangement, substitution, and result with units.

(d) [2 marks] $\displaystyle P_{\mathrm{loss}}=P_1-mcs_1=100-(0.500)(1000)(0.150)=25.0\,\mathrm{W}$ Run 2 gives the same result: $160-(0.500)(1000)(0.270)=25.0\,\mathrm{W}$ .

(e) [3 marks] For differences, add absolute uncertainties: $\displaystyle \delta(P_2-P_1)=1+1=2\,\mathrm{W}$ $\displaystyle \delta(s_2-s_1)=0.010+0.010=0.020\,\mathrm{K\,s^{-1}}$ For the quotient, add fractional contributions: $\displaystyle \frac{\delta c}{c}\approx\frac{2}{60}+\frac{0.005}{0.500}+\frac{0.020}{0.120}=0.210$ The maximum percentage uncertainty is $21\%$ , so $\displaystyle c=(1000\pm210)\,\mathrm{J\,kg^{-1}\,K^{-1}}$ The gradient difference dominates this estimate. This is a maximum linear estimate, not a quadrature calculation.

(f) [4 marks] One limitation is that losses increase with block temperature, so the common-loss assumption is only approximate. Improve this by comparing local gradients over the same narrow temperature interval in both runs, or by modelling losses as a function of temperature. A second limitation is that the heater and probe also absorb energy. Ignoring their heat capacities makes the inferred block $c$ too large: calibrate their combined heat capacity $C_{\mathrm{extra}}$ and use $c=[(P_2-P_1)/(s_2-s_1)-C_{\mathrm{extra}}]/m$ . Credit two explained limitation–improvement pairs. Other valid pairs include temperature gradients with improved contact or multiple probes, and noisy gradients with repeated runs and fitted slopes.

Exam technique checklist

Separate temperature changes from absolute temperatures; do not add 273.15 to a temperature difference.
Split a heating-and-melting calculation into stages and check whether the available energy can complete the phase change.
For calorimetry, include every object whose heat capacity matters and state the assumptions behind energy conservation.
Distinguish emitted radiative power from net radiative power; convert nanometres to metres before using Wien’s law.
For a graph, state both axes and the meaning and units of its gradient.
When estimating a maximum uncertainty, add absolute uncertainties for differences and weighted fractional uncertainties for products, quotients, and powers.
Evaluate experiments by linking each limitation to its effect and a specific improvement.

Faith connection — Responsible use of energy

Genesis 2:15 presents care for creation as a human responsibility. Understanding thermal energy transfer can help us make thoughtful choices about insulation, heating, and resource use. Reflect on how scientific understanding and responsible stewardship can work together when improving a home or school’s energy efficiency.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Thermal energy transfers HL Papers 1& 2

B.1 — Thermal Energy TransfersHL Practice: Papers 1 & 2

How to use this practice bank

Data and equations

Paper 1A style — Multiple-choice questions

Question 1 — Temperature scales [1 mark]

Question 2 — Temperature differences [1 mark]

Question 3 — Internal energy [1 mark]

Question 4 — Density [1 mark]

Question 5 — Specific heat capacity [1 mark]

Question 6 — Heating-rate comparison [1 mark]

Question 7 — Melting and particle energy [1 mark]

Question 8 — Latent heat [1 mark]

Question 9 — Heating efficiency [1 mark]

Question 10 — Thermal equilibrium [1 mark]

Question 11 — Incomplete melting [1 mark]

Question 12 — Evaporative cooling [1 mark]

Question 13 — Conduction in metals [1 mark]

Question 14 — Natural convection [1 mark]

Question 15 — Conduction and thickness [1 mark]

Question 16 — Conduction calculation [1 mark]

Question 17 — Stefan–Boltzmann scaling [1 mark]

Question 18 — Inverse-square spreading [1 mark]

Question 19 — Wien’s law [1 mark]

Question 20 — Radiative equilibrium [1 mark]

Paper 2 style — Extended-response questions

Extended response 1 — Temperature scales [5 marks]

Extended response 2 — Specific heat capacity [5 marks]

Extended response 3 — Melting and internal energy [5 marks]

Extended response 4 — Combined warming and melting [7 marks]

Extended response 5 — Temperature, internal energy, and compression [6 marks]

Extended response 6 — Kettle power and efficiency [7 marks]

Extended response 7 — Mixing water and the effect of a container [7 marks]

Extended response 8 — Calorimetry with a hot metal [7 marks]

Extended response 9 — Interpreting a heating curve [8 marks]

Extended response 10 — Ice added to warm water [8 marks]

Extended response 11 — Conduction through a wall and insulation [7 marks]

Extended response 12 — Thermal-transfer mechanisms and net radiation [8 marks]

Extended response 13 — Black-body temperature, luminosity, and brightness [7 marks]

Extended response 14 — Inferring stellar distance and uncertainty [9 marks]

Extended response 15 — Radiative cooling and changing power [8 marks]

Lab investigation — Heating with energy losses

Investigation 1 — Determine a metal’s specific heat capacity [20 marks]

Exam technique checklist

Faith connection — Responsible use of energy

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B.1 — Thermal Energy Transfers
HL Practice: Papers 1 & 2