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Questions for Thermal Physics (JC) Set 1

A Level Physics thermal physics practice questions (JC Set 1), with worked answers.

  • GCE A-Level H2 Physics 2027
Learning goals
  • Work through the questions and use the feedback to plan revision.

Learning objectives

  • Use thermodynamic temperature and convert between Celsius and kelvin.
  • Apply work conventions and the zeroth and first laws of thermodynamics.
  • Define and use heat capacity and specific heat capacity in energy balances.
  • Define and use specific latent heat in phase-change energy balances.

Use Current Practice First

Which of the following statements about the thermodynamic scale is not correct?

  1. It is also known as the Kelvin scale.
  2. It is a shifted scale from the Celsius scale by a value of 273.16.
  3. One degree change in the thermodynamic scale is equivalent to one degree change in the Celsius scale.
  4. The zero value of the thermodynamic scale is absolute zero.
Show/Hide Answer

The value should be 273.15, not 273.16. Answer: 2

A fixed mass of an ideal gas loses 2190 J of heat and expands under a constant pressure of 11 kPa from a volume of 25 x 10-3 m3 to a volume of 50 x 10-3 m3. What is the change in the internal energy of the gas?

  1. -1920 J
  2. -2470 J
  3. 1920 J
  4. 2470 J
Show/Hide Answer

Using Δ U = Q + Wₒₙ, the energy transferred by heating is Q = -2190 J. Expansion gives Wₒₙ = -pΔ V = -(11 × 10³)(50-25) × 10⁻³ = -275 J Hence Δ U = -2190-275 = -2465 J ≈ -2470 J Answer: 2.

Explain why the specific latent heat of fusion is usually less than the specific latent heat of vaporisation for a given substance.

Show/Hide Answer

During melting, the particles remain close together and many intermolecular attractions remain. During vaporisation, the particles must become widely separated, so substantially more energy is usually needed to overcome intermolecular attractions. The much larger volume increase during vaporisation can also require more work against atmospheric pressure. Hence the specific latent heat of vaporisation is usually greater than the specific latent heat of fusion for the same substance.

A flask contains 300 g of water at 80°C. Initially, the flask is at thermal equilibrium with the water. 40 g of ice at 0°C is added to the water in the flask. After some time, when all the ice had melted, the temperature of the flask and its contents was found to be 50°C. a) Neglecting any heat lost to the surroundings, show that Qflask = 0.040lf - 29400, where Qflask is the heat lost by the flask and lf is the specific latent heat of fusion of ice. (Specific heat capacity of water = 4200 J kg-1K-1 b) A further 65 g of ice was then added to the flask. After all the ice had melted, it was noted that the temperature of the flask and its contents was 20°C. Q’flask, the heat lost by the flask after 65 g of ice was added was found to be related to lf, the specific latent heat of fusion of ice by the expression Q’flask = 0.065lf - 37380 i) Explain why the heat lost by the flask is the same for both situations in a) and b). ii) Hence, find lf, the specific latent heat of fusion of ice.

Show/Hide Answer

(a)Heat gained by the ice for melting = 0.04 lf Heat gained by the melted ice to raise its temperature from 0°C to 40°C = 0.04 x 4200 x (50 - 0) Heat lost by the water in the flask = 0.3 x 4200 x (80 - 50) Heat lost by the flask = mf x cf x (80 - 50), where mf and cf is the mass and specific heat capacity of the flask respectively.Heat gained = Heat lost 0.04 lf + 0.04 x 4200 x (50 - 0) = 0.3 x 4200 x (80 - 50)+ mf x cf x (80 - 50) Qf = mf x cf x (80 - 50) - 0.3 x 4200 x (80 - 50) Qf = 0.04lf - 29400 J b) i) The temperature change is the same for both situations, hence the heat lost by the flask will be the same. ii) 0.040lf - 29400 = 0.065lf - 37380 lf = 3.19 x 105 J kg-1

In an experiment with a continuous flow calorimeter to find the specific heat capacity of a liquid, an input power of 60 W produced a rise in temperature of 10 K in the liquid. When the power was doubled, the same temperature rise was achieved by making the rate of flow of the liquid three times faster. The power lost to the surroundings in each case was

  1. 20 W
  2. 30 W
  3. 40 W
  4. 50 W
Show/Hide Answer

E = mcΔθ + heat lost Pt = mcΔθ + heat lost 60t = mcΔθ + heat lost ---- eqn 1 120t = 3mcΔθ + heat lost ---- eqn 2 (3m because rate of flow of liquid is 3 times faster) Solving eqn 1 and eqn 2, heat lost = 30W Answer: 2

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Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027