A Level Physics: Thermodynamics — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Thermodynamics revision notes

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Section A

1. Define internal energy. [2]

2. State the first law of thermodynamics and define each term. [3]

3. Explain why the temperature of a substance remains constant while it is melting. [2]

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Section B

4. A fixed mass of an ideal gas is compressed. 250 J of work is done on the gas and it loses 80 J of thermal energy to the surroundings.

(a) Determine the change in internal energy, showing the sign of each term. [3]

(b) State and explain what happens to the temperature of the gas. [2]

5. An electric heater rated at 60 W is used to melt 0.15 kg of ice at 0 °C. It takes 8.5 minutes to melt completely.

(a) Calculate the energy supplied. [2]

(b) Calculate the specific latent heat of fusion of ice from these data. [2]

(c) The accepted value is 3.34 × 10⁵ J kg⁻¹. Suggest why the experimental value is lower, and one improvement. [3]

6. 0.50 kg of water at 18 °C is heated to 100 °C and then completely boiled away. (c_water = 4180 J kg⁻¹ K⁻¹; L_v = 2.26 × 10⁶ J kg⁻¹)

(a) Calculate the energy needed to raise the temperature to 100 °C. [2]

(b) Calculate the energy needed to boil it. [2]

(c) Explain why the latent heat of vaporisation is much greater than the latent heat of fusion. [2]

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Section C

7. A fixed mass of ideal gas undergoes an isothermal expansion. A separate fixed mass of ideal gas undergoes an adiabatic compression.

(a) State what ΔU equals for the isothermal process, and hence the relationship between q and W. [2]

(b) State what q equals for the adiabatic process, and hence the relationship between ΔU and W. [2]

(c) Explain what is meant by internal energy being a state function. [2]

8. A gas at a constant pressure of 1.0 × 10⁵ Pa expands from a volume of 2.0 × 10⁻³ m³ to 3.2 × 10⁻³ m³, absorbing 620 J of thermal energy.

(a) Calculate the work done by the gas as it expands. [2]

(b) Calculate the change in internal energy, applying the correct sign convention. [3]

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Answers

1. The sum of the random distribution of kinetic and potential energies of the molecules [1] of the system [1]. Both components are needed. “The total kinetic energy” scores one at most.

2. ΔU = q + W [1], where q is the thermal energy supplied to the system [1] and W is the work done on the system [1].

3. The energy supplied increases the potential component of internal energy by separating the molecules [1], rather than the kinetic component, and temperature depends on molecular kinetic energy [1].

4. (a) W = +250 J (work done on the gas) [1]; q = −80 J (thermal energy lost) [1]. ΔU = 250 + (−80) = +170 J [1].

(b) Temperature increases [1], because internal energy has risen and the kinetic component of internal energy depends on temperature [1].

5. (a) E = Pt = 60 × (8.5 × 60) [1] = 30 600 J [1].

(b) L = E ÷ m = 30 600 ÷ 0.15 [1] = 2.04 × 10⁵ J kg⁻¹ [1].

(c) The ice and its melting water start below room temperature, so heat also flows in from the warmer surroundings in addition to the heater [1]. This background heat melts extra ice without adding to the measured electrical energy E, so the mass melted is larger than the heater’s energy alone would produce, and L = E ÷ m comes out lower than the true value [1]. Improvement: run a control experiment with the heater switched off (or disconnected) for the same time, measure the mass of ice that melts from background heat alone, and subtract this from the heater-on mass before calculating L [1].

6. (a) E = mcΔθ = 0.50 × 4180 × 82 [1] = 171 380 J ≈ 1.71 × 10⁵ J [1].

(b) E = mL = 0.50 × 2.26 × 10⁶ [1] = 1.13 × 10⁶ J [1].

(c) Boiling must completely separate the molecules, overcoming the intermolecular forces entirely [1], and also do work pushing back the atmosphere as the vapour expands; melting only loosens the forces [1].

7. (a) ΔU = 0, since temperature (and hence the kinetic component of internal energy) is unchanged [1]; therefore q = −W [1].

(b) q = 0, since no heat is transferred [1]; therefore ΔU = W [1].

(c) Internal energy depends only on the current state of the system — its temperature, for an ideal gas — never on the process or route used to reach that state [1]. Two different routes between the same start and end states give the same ΔU, even though q and W individually may differ along each route [1].

8. (a) W (by the gas) = pΔV = 1.0 × 10⁵ × (3.2 × 10⁻³ − 2.0 × 10⁻³) [1] = 120 J [1].

(b) Since the gas expands, it does work on its surroundings, so the work done on the gas is W = −120 J [1]. ΔU = q + W = 620 + (−120) [1] = 500 J [1].

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Where marks are usually lost

• Defining internal energy as kinetic energy only.
• Getting the sign of W wrong for compression versus expansion.
• Using E = mcΔθ during a change of state.
• Forgetting to convert minutes to seconds.
• Explaining the latent heat difference without mentioning work against the atmosphere.
• Describing melting or vaporisation as “breaking bonds” between molecules — there is no bond-breaking model at this level. The correct description is molecules being separated (or fully separated, for vaporisation) against the intermolecular forces of attraction, which increases their potential energy.
• Applying W = pΔV when the pressure is not constant during the process.
• Forgetting that internal energy is a state function, and treating it as dependent on the path taken between two states.