OCR A Physics · Module 5: Newtonian world and astrophysics
Thermal physics: practice questions with answers
6 exam-style questions on thermal physics. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Rearranging equations2 marks
A 2.0 kW immersion heater warms 2.0 kg of water for 90 s. The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. Assume no energy is lost. Calculate the temperature rise.
Δθ = K
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Answer Δθ = 21.5 K
Energy = Pt = 2000 × 90 = 1.8 × 10⁵ J. Δθ = Q ÷ mc = 1.8 × 10⁵ ÷ (2.0 × 4180) = 21.5 K.
- Got 0.359 K? You used the time as 1.5 (minutes). Use t = 90 s: Δθ = Pt ÷ mc = 180 000 ÷ 8360 = 21.5 K.
- Got 43.1 K? You left out the mass. Q = mcΔθ, so Δθ = 180 000 ÷ (2.0 × 4180) = 21.5 K.
- Standard form3 marks
A gas at a pressure of 1.2 × 10⁵ Pa and temperature 27 °C occupies 2.5 × 10⁻³ m³. Calculate the number of molecules in the gas.
N = molecules
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Answer N = 7.25 × 10²² molecules
T = 27 + 273 = 300 K. pV = NkT gives N = (1.2 × 10⁵ × 2.5 × 10⁻³) ÷ (1.38 × 10⁻²³ × 300) = 7.25 × 10²².
- Got 8.05 × 10²³ molecules? You used the temperature in °C. Convert to kelvin: T = 300 K, so N = pV ÷ kT = 7.25 × 10²².
- Got 7.25 × 10²⁵ molecules? You used V = 2.5 m³. The volume is 2.5 × 10⁻³ m³, so N = pV ÷ kT = 7.25 × 10²².
- Using the right equation2 marks
A sealed container of gas at 17 °C has a pressure of 180 kPa. It is heated at constant volume to 77 °C. Calculate the new pressure, in kPa.
p = kPa
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Answer p = 217 kPa
At constant volume p/T is constant, with T in kelvin: p₂ = 180 × 350 ÷ 290 = 217 kPa.
- Got 815 kPa? You used temperatures in °C. Convert to kelvin (290 K and 350 K): p₂ = 180 × 350 ÷ 290 = 217 kPa.
- Got 149 kPa? You inverted the temperature ratio. Pressure rises with temperature: p₂ = 180 × 350 ÷ 290 = 217 kPa.
- Using the right equation3 marks
The mass of a helium atom is 6.64 × 10⁻²⁷ kg. Calculate the root mean square speed of helium atoms at 300 K.
crms = m s⁻¹
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Answer crms = 1370 m s⁻¹
⟨c²⟩ = 3kT ÷ m = 3 × 1.38 × 10⁻²³ × 300 ÷ 6.64 × 10⁻²⁷, so crms = 1.37e+03 m s⁻¹.
- Got 790 m s⁻¹? You left out the 3. ½m⟨c²⟩ = (3/2)kT gives ⟨c²⟩ = 3kT ÷ m, so crms = 1.37e+03 m s⁻¹.
- Got 1.87 × 10⁶ m s⁻¹? You did not take the square root. That is ⟨c²⟩ = 1.87 × 10⁶; crms = √⟨c²⟩ = 1.37e+03 m s⁻¹.
- Using the right equation1 mark
What is the internal energy of a substance?
- A The sum of the random kinetic energies of its molecules only.
- B The energy transferred to it by heating only.
- C The sum of the random kinetic and potential energies of its molecules.
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Answer C: The sum of the random kinetic and potential energies of its molecules.
Internal energy is the total of random kinetic and potential energies of the particles.
- Why not A? Intermolecular potential energy also contributes, which matters during changes of state.
- Why not B? Heating is a transfer; internal energy is a stored property that can also change by doing work.
- Using the right equation1 mark
Why does Brownian motion of smoke particles support the kinetic model?
- A The smoke particles are hit unevenly by fast, randomly moving air molecules.
- B Air molecules are visible under the microscope.
- C The smoke particles attract each other.
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Answer A: The smoke particles are hit unevenly by fast, randomly moving air molecules.
Unbalanced random collisions with invisible air molecules cause the visible jerky motion.
- Why not B? Air molecules are far too small to see; only the smoke particles are visible.
- Why not C? The smoke particles are moved by collisions, not by attraction between them.