Q1(c). Two metal blocks in thermal contact
Two blocks of a metal having same mass have same heat capacity 15 J K⁻¹ (presumed independent of T), but are at different temperatures 300 K and 400 K, respectively. Calculate the final temperature and entropy change, if the blocks are in thermal contact and reach equilibrium. (Assume two blocks together constitute the isolated system)
Answer in brief
T_f=350 K; Δ Sₜₒₜₐₗ=+0.309 J K⁻¹ (spontaneous, irreversible)
Key points
- Final temperature. No heat leaves the system: C(T_f-300)=C(400-T_f), so T_f=350 K. Heat passed: q=15×50=750 J.
- Entropy of each block. Entropy is a state function, so each change is found along a reversible path, dS=C dT/T: Δ S=Cln(T_f/Tᵢ)
- Cold block: Δ S₁=15ln(350)/(300)=+2.312 J K⁻¹
- Hot block: Δ S₂=15ln(350)/(400)=-2.003 J K⁻¹
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Given: C=15 J K⁻¹ for each block; T₁=300 K, T₂=400 K.
- Final temperature. No heat leaves the system: C(T_f-300)=C(400-T_f), so T_f=350 K. Heat passed: q=15×50=750 J.
- Entropy of each block. Entropy is a state function, so each change is found along a reversible path, dS=C dT/T: Δ S=Cln(T_f/Tᵢ)
- Cold block: Δ S₁=15ln(350)/(300)=+2.312 J K⁻¹
- Hot block: Δ S₂=15ln(350)/(400)=-2.003 J K⁻¹
- Total: Δ S=Δ S₁+Δ S₂=+0.309J K⁻¹
Reading the answer: the surroundings are untouched, so Δ Sᵤₙᵢᵥₑᵣₛₑ>0. The 750 J enters the cold block at a lower T than it leaves the hot one, so the gain q/T outweighs the loss. The flow is spontaneous and irreversible.
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All parts of Question 1
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- 1(b) Liquid of higher surface tension: five properties5 marks
- 1(c) Two metal blocks in thermal contact5 marks
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- 1(j) Why low temperature favours adsorption of a gas5 marks