B.1 Thermal energy transfers – IB Physics exam questions with mark scheme

1Core4 marks

An electric kettle of power $2.2\ \mathrm{kW}$ heats $1.5\ \mathrm{kg}$ of water from $20^\circ$C to $100^\circ$C. The specific heat capacity of water is $4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$.

(a) Calculate the thermal energy needed to raise the water to $100^\circ$C. [2]

(b) The kettle transfers 85% of its electrical energy to the water. Determine the time taken. [2]

$Q=mc\Delta T = 1.5\times4180\times80$(1)
$Q=5.02\times10^5$ J ($5.0\times10^5$ J)(1)
Electrical energy needed $=Q/0.85$ so $t = Q/(0.85P)$(1)
$t=268$ s, i.e. 270 s to 2 s.f. (allow 268 s; allow 267 s if $5.0\times10^5$ J is used)(1)
Answer: (a) $5.0\times10^5$ J (b) 270 s
Examiner tip: A temperature change of 80 K is the same as 80°C; do not convert it incorrectly.

2Standard5 marks

The spectrum of a star, assumed to be a black body, has a peak intensity at a wavelength of $4.8\times10^{-7}\ \mathrm{m}$. The radius of the star is $7.0\times10^{8}\ \mathrm{m}$. Wien's constant is $2.90\times10^{-3}\ \mathrm{m\,K}$.

(a) Determine the surface temperature of the star. [2]

(b) Calculate the luminosity of the star. [2]

(c) A second star has the same surface temperature but twice the radius. State the ratio of its luminosity to that of the first star. [1]

$T=2.90\times10^{-3}/\lambda_{max}$ (Wien's law)(1)
$T=6042$ K ($6.0\times10^3$ K)(1)
$L=\sigma A T^4$ with $A=4\pi R^2$(1)
$L=\sigma\times4\pi R^2T^4=4.7\times10^{26}$ W (allow $4.5\times10^{26}$ W if $T=6.0\times10^3$ K is used)(1)
Ratio = 4 (luminosity ∝ $R^2$)(1)
Answer: (a) $6.0\times10^3$ K (b) $4.7\times10^{26}$ W (c) 4
Examiner tip: Use the surface area of a sphere, $4\pi R^2$, and temperature in kelvin to the fourth power.

3Core4 marks

A rectangular block of aluminium measures $5.0\ \mathrm{cm}\times4.0\ \mathrm{cm}\times10.0\ \mathrm{cm}$ and has a mass of $540\ \mathrm{g}$.

(a) Calculate the density of aluminium in $\mathrm{kg\,m^{-3}}$. [2]

At $100^\circ$C and atmospheric pressure, the density of liquid water is $958\ \mathrm{kg\,m^{-3}}$ and the density of steam is $0.598\ \mathrm{kg\,m^{-3}}$.

(b) Use the molecular model of matter to explain why the density of steam is so much smaller than the density of liquid water at the same temperature. [1]

(c) Estimate the ratio $\dfrac{\text{mean separation of molecules in steam}}{\text{mean separation of molecules in liquid water}}$. [1]

(a) $V = 5.0\times4.0\times10.0 = 200\ \mathrm{cm^3} = 2.0\times10^{-4}\ \mathrm{m^3}$ (conversion of volume and mass to SI)(1)
$\rho = \dfrac{0.540}{2.0\times10^{-4}} = 2.7\times10^{3}\ \mathrm{kg\,m^{-3}}$ (award 1 max for $2.7\ \mathrm{g\,cm^{-3}}$ unconverted)(1)
(b) same mass of molecules / same molecules occupy a much larger volume in the gas because the molecules are much further apart (in a liquid they are close together, almost touching); do not accept 'gas molecules are smaller/lighter'(1)
(c) volume per molecule is inversely proportional to density, so volume ratio $=\dfrac{958}{0.598}\approx1.6\times10^{3}$; separation ratio $=\sqrt[3]{1600}\approx12$ (accept 11 to 12; do not accept 1600)(1)
Answer: (a) $2.7\times10^3\ \mathrm{kg\,m^{-3}}$ (c) about 12
Examiner tip: Separation depends on the cube root of the volume ratio: quoting 1600 for the spacing ratio loses the mark.

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4Core5 marks

A sealed container holds a mixture of helium gas and argon gas in thermal equilibrium at a temperature of $27^\circ$C. The molar mass of helium is $4.0\ \mathrm{g\,mol^{-1}}$ and that of argon is $40\ \mathrm{g\,mol^{-1}}$.

(a) Calculate the average kinetic energy of a helium atom. [2]

(b) Determine the temperature, in $^\circ$C, at which the average kinetic energy of a helium atom would be twice your answer to (a). [1]

(c) Compare the average kinetic energy and the average speed of an argon atom with those of a helium atom in the container at $27^\circ$C. [2]

(a) $T = 27+273 = 300\ \mathrm{K}$ used in $\bar E_k=\tfrac32 k_BT$(1)
$\bar E_k = 1.5\times1.38\times10^{-23}\times300 = 6.2\times10^{-21}\ \mathrm{J}$ (award 0 for use of 27 K: $5.6\times10^{-22}$ J)(1)
(b) $T = 600\ \mathrm{K} = 327^\circ$C (do not accept $54^\circ$C)(1)
(c) average kinetic energy is the same for both, because they are at the same temperature(1)
argon atoms are 10 times more massive so their average speed is smaller; (rms) speed is $\sqrt{10}\approx3.2$ times smaller (accept 'smaller' alone for the mark)(1)
Answer: (a) $6.2\times10^{-21}$ J (b) $327^\circ$C (c) same mean KE; argon slower by a factor of about 3.2
Examiner tip: Doubling the Celsius temperature does not double the kinetic energy: always double the kelvin temperature.

5Standard5 marks

A $0.25\ \mathrm{kg}$ sample of a solid is heated in an insulated container by an immersed heater that supplies energy at a constant rate of $60\ \mathrm{W}$. The graph shows how the temperature of the sample varies with time. Assume all the energy from the heater is transferred to the sample.

(a) Describe the energy changes of the molecules of the sample between $t = 200\ \mathrm{s}$ and $t = 800\ \mathrm{s}$. [2]

(b) Determine the specific latent heat of fusion of the substance. [2]

(c) Determine the ratio $\dfrac{\text{specific heat capacity of the liquid}}{\text{specific heat capacity of the solid}}$. [1]

(a) the substance is melting; the (random) kinetic energy of the molecules stays constant / temperature constant(1)
the energy supplied increases the (intermolecular) potential energy of the molecules as bonds are broken / molecules move further apart; so internal energy increases(1)
(b) $Q = Pt = 60\times600 = 3.6\times10^{4}\ \mathrm{J}$ (time read from graph as 600 s)(1)
$L = \dfrac{3.6\times10^4}{0.25} = 1.4\times10^{5}\ \mathrm{J\,kg^{-1}}$ (accept $1.44\times10^5$)(1)
(c) $c = \dfrac{P}{m(\Delta T/\Delta t)}$ so ratio = gradient(solid)/gradient(liquid) $= \dfrac{40/200}{30/250} = 1.7$ (accept $c_s=1200$, $c_l=2000\ \mathrm{J\,kg^{-1}\,K^{-1}}$ leading to 1.67)(1)
Answer: (b) $1.4\times10^5\ \mathrm{J\,kg^{-1}}$ (c) 1.7
Examiner tip: A shallower gradient means a larger specific heat capacity: students often invert the ratio in (c).

6Standard5 marks

A copper rod of length $0.30\ \mathrm{m}$ and cross-sectional area $3.0\times10^{-4}\ \mathrm{m^2}$ is surrounded by insulating lagging. One end is in boiling water at $100^\circ$C and the other end is in a mixture of ice and water at $0^\circ$C, as shown. The thermal conductivity of copper is $390\ \mathrm{W\,m^{-1}\,K^{-1}}$ and the specific latent heat of fusion of ice is $3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$.

(a) Calculate the rate of thermal energy transfer along the rod. [2]

(b) Determine the mass of ice that melts in $5.0$ minutes because of the energy conducted along the rod. [2]

(a) $\dfrac{\Delta Q}{\Delta t} = kA\dfrac{\Delta T}{\Delta x} = \dfrac{390\times3.0\times10^{-4}\times100}{0.30}$(1)
$= 39\ \mathrm{W}$(1)
(b) $Q = 39\times300 = 1.17\times10^{4}\ \mathrm{J}$ (time in seconds)(1)
$m = \dfrac{1.17\times10^4}{3.34\times10^5} = 0.035\ \mathrm{kg}$ (35 g) (allow ECF from (a))(1)
(c) without lagging energy would be lost from the sides of the rod (by convection/radiation), so the rate of transfer would not be the same along the whole rod / the temperature gradient would not be uniform(1)
Answer: (a) 39 W (b) 0.035 kg
Examiner tip: Remember $\Delta T$ is the temperature difference across the rod (100 K) and $\Delta x$ is the rod length in metres.

7Standard6 marks

A pan containing water at $20^\circ$C is heated steadily until the water boils at $100^\circ$C and then continues to boil away.

Explain, in terms of the molecules of the water, the changes in internal energy that occur during this process. Your answer should explain why the temperature stays constant while the water boils, and why the specific latent heat of vaporisation of water is much larger than its specific latent heat of fusion. [6]

Internal energy is the total random kinetic energy plus the total intermolecular potential energy of the molecules(1)
From 20 to 100 °C the energy supplied mainly increases the mean (random) kinetic energy of the molecules, so temperature rises(1)
Temperature is a measure of the average kinetic energy of the molecules ($\bar E_k=\tfrac32k_BT$)(1)
During boiling the energy supplied does work against intermolecular forces / breaks intermolecular bonds, increasing the potential energy(1)
Mean kinetic energy stays constant during boiling, so the temperature stays constant even though internal energy increases(1)
Vaporisation requires molecules to be separated completely (to large distances where forces are negligible), whereas melting only loosens the bonds and molecules remain close together(1)
So the increase in potential energy per kilogram (and the work done pushing back the atmosphere as the volume increases greatly) is much larger for vaporisation than for fusion(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, linked explanation: internal energy defined as KE + PE; rise in temperature linked to increasing mean KE; constant temperature during boiling explained by energy going into PE with mean KE constant; L_v > L_f explained by complete separation of molecules compared with loosening in melting. Correct terminology throughout.
Level 2	3–4	Explains either the heating stage or the boiling stage correctly in terms of kinetic and potential energy, with some attempt at the latent-heat comparison; some links missing or minor confusions (e.g. 'heat' for internal energy).
Level 1	1–2	Isolated relevant statements, e.g. molecules move faster when heated or bonds break when boiling, with no clear distinction between kinetic and potential energy.
Level 0	0	No relevant content.

Answer: Heating increases mean KE (temperature rises); at boiling the energy increases intermolecular PE only, so T is constant; vaporisation separates molecules completely, so needs far more energy per kg than melting.
Examiner tip: Do not say the energy 'is used to break bonds' without stating that this increases potential energy while kinetic energy stays constant.

8Hard6 marks

A student determines the specific latent heat of vaporisation $L$ of water. An electric heater is placed in water in a beaker standing on a balance. Once the water is boiling steadily, the student records the mass of water boiled away in $300\ \mathrm{s}$. The experiment is repeated with a different heater power.

Heater power / W	Mass boiled away in 300 s / g
40.0	3.6
70.0	7.6

(a) Explain why the student waits until the water is boiling steadily before starting the timing. [1]

(b) Energy is lost to the surroundings at a rate $h$. Explain why using $P_2 - P_1 = \dfrac{L(m_2-m_1)}{t}$ removes the effect of this loss. [1]

(d) Calculate $h$. [1]

(e) Suggest one reason why your value of $L$ may still be larger than the true value. [1]

(a) so that all the energy from the heater is used to vaporise water (none used to raise the temperature) / temperature is constant at the boiling point / rate of loss to surroundings is steady(1)
(b) $P_1 t = m_1L + ht$ and $P_2 t = m_2L + ht$; subtracting removes $ht$ because the water is at the same temperature (100 °C) in both runs so losses are equal(1)
(c) $L = \dfrac{(70.0-40.0)\times300}{(7.6-3.6)\times10^{-3}}$ (mass converted to kg)(1)
$L = 2.3\times10^{6}\ \mathrm{J\,kg^{-1}}$ (accept $2.25\times10^{6}$)(1)
(d) $h = 40.0 - \dfrac{2.25\times10^{6}\times3.6\times10^{-3}}{300} = 13\ \mathrm{W}$ (allow ECF from (c))(1)
(e) rate of energy loss is greater at the higher power (e.g. heater surface hotter, more vigorous boiling), so $h$ is not constant and some of $P_2-P_1$ is not used for vaporisation; accept: steam condenses on the heater/lid and drips back, so the mass boiled away is under-recorded (more at higher power). Do not accept 'heat loss' alone without why it does not cancel(1)
Answer: (c) $2.3\times10^6\ \mathrm{J\,kg^{-1}}$ (d) 13 W
Examiner tip: Convert grams to kilograms before dividing: forgetting gives an answer 1000 times too small.

9Hard5 marks

Two stars, A and B, are assumed to radiate as black bodies. The peak wavelength in the spectrum of A is $350\ \mathrm{nm}$ and that of B is $700\ \mathrm{nm}$. Star A is three times as far from Earth as star B, and the apparent brightness of A is four times that of B.

(a) Calculate the surface temperature of star A. [1]

(b) Determine the ratio $\dfrac{L_A}{L_B}$ of the luminosities of the stars. [2]

(a) $T_A = \dfrac{2.9\times10^{-3}}{350\times10^{-9}} = 8.3\times10^{3}\ \mathrm{K}$(1)
(b) $L = 4\pi d^2 b$ so $\dfrac{L_A}{L_B} = \dfrac{b_A}{b_B}\left(\dfrac{d_A}{d_B}\right)^2$(1)
$= 4\times3^2 = 36$(1)
(c) $T_A/T_B = 2$ (from Wien's law) and $L=\sigma 4\pi R^2T^4$ so $\dfrac{R_A}{R_B} = \sqrt{\dfrac{L_A/L_B}{(T_A/T_B)^4}} = \sqrt{\dfrac{36}{16}}$(1)
$= 1.5$ (allow ECF from (b))(1)
Answer: (a) $8.3\times10^3$ K (b) 36 (c) 1.5
Examiner tip: Temperature enters to the fourth power and radius is squared: forgetting the square root in (c) gives 2.25.

10Stretch5 marks

An ice cube of mass $0.040\ \mathrm{kg}$ at $-18^\circ$C is dropped into $0.30\ \mathrm{kg}$ of water at $22^\circ$C in an aluminium cup of mass $0.12\ \mathrm{kg}$, also at $22^\circ$C. Energy exchange with the surroundings is negligible.

specific heat capacity of ice	$2100\ \mathrm{J\,kg^{-1}\,K^{-1}}$
specific heat capacity of water	$4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$
specific heat capacity of aluminium	$900\ \mathrm{J\,kg^{-1}\,K^{-1}}$
specific latent heat of fusion of ice	$3.34\times10^{5}\ \mathrm{J\,kg^{-1}}$

(a) Show that all of the ice melts. [2]

(b) Determine the final temperature of the contents of the cup. [3]

(a) energy needed to warm ice to 0 °C and melt it $= 0.040\times2100\times18 + 0.040\times3.34\times10^5 = 1512 + 13360 \approx 1.49\times10^{4}\ \mathrm{J}$(1)
energy released if water and cup cool to 0 °C $= (0.30\times4180 + 0.12\times900)\times22 \approx 3.00\times10^{4}\ \mathrm{J}$, which is greater, so all the ice melts (comparison needed)(1)
(b) energy gained = energy lost with the melted ice warming from 0 °C: $1.49\times10^4 + 0.040\times4180\,T = 1362\,(22 - T)$(1)
correct rearrangement: $T = \dfrac{29964 - 14872}{1362 + 167.2}$(1)
$T = 9.9^\circ$C (accept 9.8–9.9; award 2 max for 8.9 °C from omitting the cup)(1)
Answer: Final temperature $9.9^\circ$C
Examiner tip: The melted ice must then be warmed from 0°C as water: forgetting this term or the cup are the commonest errors.

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