Theme B · The particulate nature of matter

A 2.0 kW heater supplies energy to 0.30 kg of ice at 0 °C until it has all melted. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg⁻¹. How long does this take, assuming no losses?

1. A5.0 s
2. B50 s
3. C165 s
4. D500 s

The solar constant is 1360 W m⁻² and the Earth's average albedo is 0.30. What is the average intensity of solar radiation absorbed per square metre of the Earth's surface?

1. A238 W m⁻²
2. B340 W m⁻²
3. C952 W m⁻²
4. D1360 W m⁻²

A fixed mass of ideal gas is at 27 °C. It is heated at constant volume until its pressure has doubled. What is the new temperature?

1. A54 °C
2. B327 °C
3. C600 °C
4. D873 °C

A wire of resistivity ρ, length L and cross-sectional area A has resistance R. A second wire of the same material has twice the length and twice the diameter. What is its resistance?

1. AR/2
2. BR
3. C2R
4. D4R

An ideal gas is compressed adiabatically. Which statement is correct?

1. ANo thermal energy enters or leaves the gas, and its temperature falls.
2. BNo thermal energy enters or leaves the gas, and its temperature rises.
3. CThermal energy leaves the gas, and its temperature stays constant.
4. DThermal energy enters the gas, and its temperature rises.

Which method of thermal energy transfer can occur through a vacuum?

1. Aconduction only
2. Bconvection only
3. Cradiation only
4. Dall three

A student determines the specific heat capacity of a liquid. She places 0.500 kg of the liquid in a well-insulated container with an electric heater of constant power 50.0 W and records the temperature at intervals.

1. (a)

   Determine Determine the gradient of a graph of temperature against time, and explain why the gradient rather than a single pair of readings should be used.

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   1. 1

      gradient = (31.4 − 20.0)/480 = 0.0238 °C s⁻¹

   2. 2

      the gradient uses all five readings, so random errors in individual temperature measurements partly cancel

   3. 3

      and a straight line confirms that the rate of heating is constant, which the calculation assumes

2. (b)

   Determine Determine the specific heat capacity of the liquid.

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   1. 1

      Constant heater power satisfies P = mc(dθ/dt), where dθ/dt is the measured temperature gradient.

   2. 2

      c = P / (m × gradient) = 50.0 / (0.500 × 0.0238)

   3. 3

      c = 4.2 × 10³ J kg⁻¹ K⁻¹

3. (c)

   Outline Outline why the container being well insulated matters more at the end of the experiment than at the beginning.

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   1. 1

      the rate of energy transfer to the surroundings depends on the temperature difference between the liquid and the room

   2. 2

      at the start the liquid is near room temperature and loses almost nothing; by the end it is 11 °C above the room and the losses are largest

4. (d)

   Suggest The heater and the container itself also absorb energy. Suggest how this affects the value obtained for c, and suggest one way of accounting for it.

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   1. 1

      some of the 50.0 W raises the temperature of the heater and container rather than the liquid

   2. 2

      the liquid therefore warms more slowly than the model assumes, the gradient is smaller, and the calculated value of c is too large

   3. 3

      account for it by repeating the experiment with a different mass of liquid and using the difference, or by determining the thermal capacity of the container separately and subtracting it

A sealed rigid container of volume 5.0 × 10⁻³ m³ holds nitrogen gas at a pressure of 2.4 × 10⁵ Pa and a temperature of 290 K. Nitrogen may be treated as an ideal gas.

1. (a)

   Calculate Calculate the amount of gas in the container, in moles.

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   1. 1

      n = pV/RT = (2.4 × 10⁵ × 5.0 × 10⁻³)/(8.31 × 290)

   2. 2

      n = 0.50 mol

2. (b)

   Determine Determine the average kinetic energy of a nitrogen molecule in the container.

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   1. 1

      E_k = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 290

   2. 2

      E_k = 6.0 × 10⁻²¹ J

3. (c)

   Explain The container is warmed to 350 K. Explain, in terms of the behaviour of the molecules, why the pressure increases.

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   1. 1

      the average kinetic energy of the molecules increases, so their mean speed increases

   2. 2

      each molecule therefore strikes the walls more frequently

   3. 3

      and each collision produces a larger change of momentum, so the average force per unit area on the walls rises

4. (d)

   Explain Outline two assumptions of the ideal gas model, and explain why real nitrogen behaves less like an ideal gas at very high pressure.

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   1. 1

      the molecules have negligible volume compared with the volume of the container

   2. 2

      there are no intermolecular forces except during collisions, and all collisions are elastic

   3. 3

      at very high pressure the molecules are pushed close together, so their own volume is no longer negligible compared with the space available

   4. 4

      and at those separations the intermolecular attractions are significant, so the molecules exert less force on the walls than the model predicts

The Earth may be modelled as a black body of average surface temperature 288 K. The Stefan–Boltzmann constant is 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

1. (a)

   Calculate Calculate the power radiated per square metre by the Earth's surface at this temperature.

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   1. 1

      I = σT⁴ = 5.67 × 10⁻⁸ × 288⁴

   2. 2

      I = 390 W m⁻²

2. (b)

   Determine The average intensity of solar radiation absorbed by the Earth is 238 W m⁻². Determine the temperature the Earth's surface would have if it radiated only this absorbed power directly to space.

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   1. 1

      sets σT⁴ = 238

   2. 2

      T⁴ = 238/5.67 × 10⁻⁸ = 4.20 × 10⁹

   3. 3

      T = 255 K

3. (c)

   Explain Explain, using your answers to (a) and (b), how the greenhouse effect accounts for the difference between 255 K and the observed 288 K.

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   1. 1

      the Earth's surface radiates in the infrared, at wavelengths determined by its temperature

   2. 2

      greenhouse gases in the atmosphere absorb strongly in the infrared while remaining largely transparent to the incoming shorter-wavelength solar radiation

   3. 3

      the absorbed energy is re-emitted in all directions, so a substantial part returns to the surface

   4. 4

      the surface therefore receives energy from the Sun and from the atmosphere, and must warm to 288 K before it radiates 390 W m⁻² and reaches equilibrium

4. (d)

   Discuss A student says that because the Earth's surface radiates 390 W m⁻² but absorbs only 238 W m⁻² from the Sun, energy is being created. Discuss this claim.

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   1. 1

      the claim is wrong: it counts only one of the two sources of energy arriving at the surface

   2. 2

      the surface also receives roughly 150 W m⁻² of infrared radiation emitted downward by the atmosphere, so the total input balances the 390 W m⁻² output

   3. 3

      the atmosphere in turn radiates to space, and it is the top of the atmosphere — not the surface — where the 238 W m⁻² balance with space must hold

Part 1. A cell of e.m.f. 6.0 V and internal resistance 0.75 Ω is connected to an external resistor of resistance 3.0 Ω. Part 2. A fixed mass of ideal monatomic gas is taken from state X to state Y by two different routes: an isothermal expansion, and an adiabatic expansion to the same final volume.

1. (a)

   Calculate Calculate the current in the circuit and the terminal potential difference of the cell.

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   1. 1

      I = ε/(R + r) = 6.0/3.75 = 1.6 A

   2. 2

      V = IR = 1.6 × 3.0

   3. 3

      V = 4.8 V

2. (b)

   Determine Determine the percentage of the power produced by the cell that is dissipated inside the cell itself.

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   1. 1

      power in r = I²r = 1.6² × 0.75 = 1.92 W

   2. 2

      total power = εI = 6.0 × 1.6 = 9.6 W

   3. 3

      percentage = 1.92/9.6 = 20%

3. (c)

   Compare Compare the final temperature and the work done by the gas for the isothermal and the adiabatic expansions to the same final volume.

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   1. 1

      isothermal: the temperature is unchanged by definition, so ΔU = 0 and the work done by the gas equals the thermal energy absorbed

   2. 2

      adiabatic: no thermal energy enters, so the work done by the gas comes entirely from its internal energy

   3. 3

      the adiabatic expansion therefore ends at a lower temperature than the isothermal one

   4. 4

      and because its pressure falls faster with volume, the area under its curve is smaller — less work is done by the gas

4. (d)

   Explain Explain why the entropy of the gas increases during the isothermal expansion but is unchanged during a reversible adiabatic expansion.

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   1. 1

      entropy change for a reversible process is ΔS = Q/T

   2. 2

      in the isothermal expansion the gas absorbs thermal energy at constant temperature, so Q is positive and ΔS is positive — and in molecular terms the same molecules now occupy a larger volume, so there are more available microstates

   3. 3

      in a reversible adiabatic expansion Q = 0, so ΔS = 0: the increase in volume is exactly offset by the narrowing of the molecular speed distribution as the gas cools