A2 · Paper 4 practice · Temperature

Energy in, temperature or state out.

A full set of ten original structured questions in the style of Paper 4, covering the whole of Temperature: thermal equilibrium, the thermodynamic scale, and specific heat and latent heat. Each question is linked to the lesson it draws on; attempt them all, then reveal the worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level Paper 4 style. They are not from past papers. They test the same concepts and skills the syllabus rewards. Use the data stated in each question, and T / K = θ / °C + 273.15.
Keep these straight

Heat, and what it changes.

01
Paper 4
[12 marks]

Two metal blocks A and B are placed in thermal contact and isolated from their surroundings, as shown in Fig. 1.1. Block A starts at 85 °C and block B starts at 25 °C.

A 85 °C B 25 °C energy two blocks in thermal contact, isolated from the surroundings
Fig. 1.1

(a) State what is meant by thermal equilibrium between two bodies. [2]

(b) For the two blocks in contact:

(i) state the direction of the net transfer of thermal energy and explain your answer, [3]

(ii) state what is true of the two temperatures once thermal equilibrium is reached. [1]

(c) Using T / K = θ / °C + 273.15:

(i) express the starting temperature of block A in kelvin, [1]

(ii) a block later has a thermodynamic temperature of 300 K; state this in °C. [1]

(d) State the Celsius value of absolute zero, and state what is special about it in terms of the energy of the particles. [2]

(e) State one reason the thermodynamic temperature scale is preferred to a scale defined by the expansion of a particular liquid. [2]

  • (a)Two bodies are in thermal equilibrium when there is no net transfer of thermal energy between them ✓, which happens when they are at the same temperature ✓
  • (b)(i)Net energy is transferred from A to B ✓, because A is at the higher temperature ✓; net flow is always from higher to lower temperature ✓
  • (b)(ii)The two temperatures become equal ✓
  • (c)(i)T = 85 + 273.15 = 358 K ✓
  • (c)(ii)θ = 300 − 273.15 = 27 °C ✓
  • (d)Absolute zero is −273 °C (−273.15 °C) ✓; it is the temperature at which the particles have their minimum (least) kinetic energy ✓
  • (e)The thermodynamic scale does not depend on the property or substance used ✓; different liquids would not agree at points between the fixed points, whereas the thermodynamic scale gives one substance-independent value ✓
02
Paper 4
[13 marks]

A metal block of mass 0.80 kg is heated by a well-lagged electric heater of power 50 W, as shown in Fig. 2.1. In a time of 4.0 minutes the temperature of the block rises from 18 °C to 52 °C.

metal block 0.80 kg heater, 50 W supply block heated electrically and well lagged
Fig. 2.1

(a) Define the specific heat capacity of a substance. [2]

(b) Calculate the electrical energy supplied by the heater in this time. [2]

(c) Assuming all this energy is transferred to the block, calculate the specific heat capacity of the metal. [3]

(d) In practice some energy is lost to the surroundings. Explain why this makes the experimental value of the specific heat capacity larger than the true value. [2]

(e) Suggest one change to the procedure that would reduce this error. [1]

(f) The same quantity of energy from part (b) is supplied instead to 0.80 kg of water, of specific heat capacity 4200 J kg⁻¹ K⁻¹. Calculate the rise in temperature and comment on its size. [3]

  • (a)The energy required per unit mass ✓ to raise the temperature of the substance by one kelvin (one degree) ✓
  • (b)E = Pt = 50 × (4.0 × 60) = 50 × 240 ✓ = 1.2 × 10⁴ J ✓
  • (c)Δθ = 52 − 18 = 34 K ✓; c = E / (mΔθ) = 1.2 × 10⁴ / (0.80 × 34) ✓ = 4.4 × 10² J kg⁻¹ K⁻¹ ✓
  • (d)Part of the supplied energy is lost to the surroundings rather than heating the block ✓; the calculation credits all of it to the block, so the energy assigned per degree of rise is too large, giving c too high ✓
  • (e)Improve the lagging / insulation (or apply a cooling correction, or use a lower power over a longer time) ✓
  • (f)Δθ = E / (mc) = 1.2 × 10⁴ / (0.80 × 4200) ✓ = 3.6 K ✓; the rise is much smaller because water has a much greater specific heat capacity than the metal ✓
03
Paper 4
[13 marks]

This question is about changes of state. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹, specific heat capacity of water = 4200 J kg⁻¹ K⁻¹, specific latent heat of fusion of ice = 3.3 × 10⁵ J kg⁻¹.

energy θ/°C −10 0 20 melting at 0 °C ice at −10 °C heated to water at 20 °C
Fig. 3.1

(a) Define the specific latent heat of fusion of a substance. [2]

(b) An electric heater of power 40 W melts 72 g of ice, already at 0 °C, in 5.0 minutes. Use this to calculate the specific latent heat of fusion of ice. [3]

(c) Explain why, while the ice is melting, energy is supplied yet the temperature does not change. [2]

(d) The specific latent heat of vaporisation of water is much greater than its specific latent heat of fusion. Suggest, in molecular terms, why this is so. [2]

(e) A mass of 0.050 kg of ice at −10 °C is heated until it becomes water at 20 °C (Fig. 3.1). Calculate the total energy required. [4]

  • (a)The energy required per unit mass ✓ to change the state from solid to liquid without any change of temperature ✓
  • (b)E = Pt = 40 × 300 = 1.2 × 10⁴ J ✓; L = E / m = 1.2 × 10⁴ / 0.072 ✓ = 1.7 × 10⁵ J kg⁻¹ ✓
  • (c)The energy is used to break or loosen the bonds between molecules, increasing their potential energy ✓; the mean kinetic energy, and so the temperature, does not change until melting is complete ✓
  • (d)In vaporisation the molecules are separated completely to large distances against the intermolecular forces (and work is done pushing back the atmosphere) ✓, whereas fusion only loosens the lattice, so vaporisation needs much more energy ✓
  • (e)Warm ice: mcΔθ = 0.050 × 2100 × 10 = 1050 J ✓; melt: mL = 0.050 × 3.3 × 10⁵ = 16500 J ✓; warm water: 0.050 × 4200 × 20 = 4200 J ✓; total = 1050 + 16500 + 4200 = 2.2 × 10⁴ J ✓
04
Paper 4
[10 marks]

Two blocks at different temperatures are placed in contact inside a thermally isolated container, as shown in Fig. 4.1.

hot T₁ cold T₂ net energy flow in contact, thermally isolated
Fig. 4.1

(a) State what is meant by thermal equilibrium. [2]

(b) State the direction of the net flow of thermal energy before equilibrium. [1]

(c) Explain, in terms of energy, why the hot block cools while the cold block warms. [2]

(d) State the quantity that determines the direction of net thermal energy flow. [1]

(e) At equilibrium, state whether the two blocks are at the same temperature and whether they hold the same internal energy. [2]

(f) Explain why two objects with equal internal energies need not be at the same temperature. [2]

  • (a)A state in which there is no net flow of thermal energy between the objects ✓, because they are at the same temperature ✓
  • (b)From the hotter block to the cooler block ✓
  • (c)There is a net transfer of energy from the hot block to the cold block ✓, lowering the internal energy of one and raising the other until temperatures are equal ✓
  • (d)The temperature (difference), not the amount of internal energy ✓
  • (e)They reach the same temperature ✓; their internal energies are generally not equal ✓
  • (f)Internal energy depends on the mass and material as well as the temperature ✓✓
05
Paper 4
[10 marks]

This question is about the thermodynamic (kelvin) temperature scale.

(a) State what is meant by absolute zero. [2]

(b) Write the relation between thermodynamic temperature T in kelvin and Celsius temperature θ. [1]

(c) Convert 27 °C to kelvin, and 200 K to degrees Celsius. [2]

(d) Explain why a temperature change of 1 °C is equal to a change of 1 K. [1]

(e) State what is meant by the thermodynamic scale being independent of the substance used. [2]

(f) The mean kinetic energy of gas molecules is proportional to T in kelvin. Explain why this relation would fail if θ in °C were used. [2]

  • (a)The lowest possible temperature, the zero of the thermodynamic scale, at which particles have minimum energy ✓✓
  • (b)T / K = θ / °C + 273.15 ✓
  • (c)27 °C = 300 K ✓; 200 K = −73 °C ✓
  • (d)The two scales have divisions of the same size ✓
  • (e)It does not depend on a property of any particular material, unlike an empirical scale ✓✓
  • (f)The proportionality holds only from absolute zero ✓; the Celsius zero is arbitrary (the ice point), so KE would not be proportional to θ ✓
06
Paper 4
[10 marks]

Thermometers measure temperature through a physical property that varies with it.

(a) State two physical properties that can be used to measure temperature. [2]

(b) Explain what is meant by calibrating a thermometer using two fixed points. [2]

(c) Explain why two different thermometers may disagree at temperatures between the fixed points. [2]

(d) State one advantage of the thermodynamic scale over an empirical scale. [2]

(e) A resistance thermometer reads 100 Ω at 0 °C and 138 Ω at 100 °C. Assuming a linear response, find the temperature when its resistance is 119 Ω. [2]

  • (a)Any two of: the volume of a liquid, electrical resistance, the e.m.f. of a thermocouple, the pressure of a fixed-volume gas ✓✓
  • (b)Marking the readings at two known temperatures (such as the ice and steam points) ✓ and dividing the scale evenly between them ✓
  • (c)Their chosen properties vary differently (non-linearly) with temperature ✓, so the empirical scales agree only at the fixed points ✓
  • (d)It is independent of the substance and agrees everywhere ✓✓
  • (e)θ = (119 − 100)/(138 − 100) × 100 = 19/38 × 100 ✓ = 50 °C ✓
07
Paper 4
[11 marks]

A 2.0 kW electric heater warms 1.5 kg of water. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.

(a) Define specific heat capacity. [2]

(b) Calculate the energy needed to raise the water from 20 °C to 100 °C. [2]

(c) Calculate the time this would take if no energy were lost. [2]

(d) In practice the heating takes longer. State and explain why. [2]

(e) Calculate the initial rate of rise of temperature of the water. [2]

(f) State one assumption made in part (e). [1]

  • (a)The energy required to raise the temperature of unit mass of a substance by one kelvin (or one degree Celsius) ✓✓
  • (b)Q = mcΔθ = 1.5 × 4200 × 80 ✓ = 5.0 × 10⁵ J ✓
  • (c)t = Q/P = 504000 / 2000 ✓ = 2.5 × 10² s ✓
  • (d)Energy is lost to the container and surroundings ✓, so more than 504 kJ must be supplied, taking longer ✓
  • (e)dθ/dt = P/(mc) = 2000 / (1.5 × 4200) ✓ = 0.32 K s⁻¹ ✓
  • (f)All the power heats the water, with no losses ✓
08
Paper 4
[11 marks]

A 0.80 kg metal block is heated by a 50 W heater for 4.0 minutes, and its temperature rises by 25 °C.

(a) Calculate the electrical energy supplied. [2]

(b) Assuming no energy is lost, calculate the specific heat capacity of the metal. [2]

(c) The accepted value is lower than this. State and explain why this method overestimates the specific heat capacity. [2]

(d) Suggest one improvement that would reduce this error. [1]

(e) Using the value from (b), calculate the initial rate of rise of temperature. [2]

(f) Explain why measuring the temperature rise over a short interval improves the accuracy. [2]

  • (a)E = Pt = 50 × (4.0 × 60) ✓ = 1.2 × 10⁴ J ✓
  • (b)c = E/(mΔθ) = 12000 / (0.80 × 25) ✓ = 6.0 × 10² J kg⁻¹ K⁻¹ ✓
  • (c)Some energy is lost to the surroundings, but all of it is attributed to the block ✓, so the calculated c is larger than the true value ✓
  • (d)Insulate (lag) the block, or correct for the heat loss ✓
  • (e)dθ/dt = P/(mc) = 50 / (0.80 × 600) ✓ = 0.10 K s⁻¹ ✓
  • (f)Less time means less energy is lost to the surroundings, so the result is closer to ideal ✓✓
09
Paper 4
[11 marks]

A 0.30 kg block of ice at 0 °C is heated until it becomes water at 40 °C. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg⁻¹ and the specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. Fig. 9.1 shows how the temperature varies with energy supplied.

θ/°C energy supplied 0 40 melting (constant T) water warming
Fig. 9.1

(a) Define the specific latent heat of fusion. [2]

(b) Calculate the energy needed to melt the ice. [2]

(c) Calculate the energy needed to warm the resulting water from 0 °C to 40 °C. [2]

(d) Calculate the total energy supplied. [1]

(e) Explain why the temperature stays constant while the ice melts, even though energy is supplied. [2]

(f) State why the specific latent heat of vaporisation is much larger than that of fusion. [2]

  • (a)The energy required to change unit mass of a substance from solid to liquid with no change in temperature ✓✓
  • (b)Q = mL = 0.30 × 3.3 × 10⁵ ✓ = 9.9 × 10⁴ J ✓
  • (c)Q = mcΔθ = 0.30 × 4200 × 40 ✓ = 5.0 × 10⁴ J ✓
  • (d)Total = 99000 + 50400 = 1.5 × 10⁵ J ✓
  • (e)The energy separates molecules (increases their potential energy) rather than their kinetic energy ✓, so the temperature does not change ✓
  • (f)Vaporisation fully separates the molecules and does work pushing back the atmosphere ✓, a far larger change than the small loosening during melting ✓
10
Paper 4
[11 marks]

A 0.50 kg copper block at 200 °C is dropped into 0.40 kg of water at 20 °C in an insulated container. This question links thermal equilibrium with specific heat capacity. Take c(copper) = 390 J kg⁻¹ K⁻¹ and c(water) = 4200 J kg⁻¹ K⁻¹.

(a) State the principle used to find the final temperature. [2]

(b) Write the energy-balance equation for the block and the water. [2]

(c) Calculate the final equilibrium temperature. [3]

(d) State the assumption made in this calculation. [1]

(e) Explain, in terms of thermal equilibrium, why the block and water end at the same temperature. [2]

(f) State whether the block and the water hold equal internal energies at equilibrium. [1]

  • (a)Conservation of energy: the thermal energy lost by the copper equals that gained by the water ✓✓
  • (b)m(Cu)c(Cu)(200 − T) = m(w)c(w)(T − 20) ✓✓
  • (c)0.50 × 390 × (200 − T) = 0.40 × 4200 × (T − 20) ✓; 195(200 − T) = 1680(T − 20) ✓; T = 39 °C ✓
  • (d)No energy is lost to the container or surroundings ✓
  • (e)Energy flows from the hotter copper to the cooler water until their temperatures are equal, when there is no net flow ✓✓
  • (f)No: they share a temperature, not an internal energy ✓

Mark this once you have attempted all ten questions and checked your working against the solutions. Revealing the solutions alone does not count.