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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2020 · Paper I

Q2(b). Heat engine between 500 K and 300 K: is 450 J from 1000 J possible?

  • 10 marks
  • Answered
Question

A person claims that a heat engine exchanges heat with reservoirs at 300 and 500 K, produces 450 J of work per 1000 J of heat extracted from the hot reservoir. Is his/her claim feasible? Give reason(s) to support your answer.

Answer in brief

Not feasible. The Carnot (maximum) efficiency is 1-300/500=0.40, so at most 400 J of work can come from 1000 J; the claim (45 %) would make the entropy of the universe fall by 0.167 J K⁻¹, which the second law forbids.

Key points

  • Claimed efficiency: η=(w)/(qₕ)=(450)/(1000)=0.45
  • no engine working between two reservoirs can be more efficient than a reversible (Carnot) engine between the same reservoirs (Carnot theorem): ηₘₐₓ=1-(T_c)/(Tₕ)=1-(300)/(500)=0.40, so wₘₐₓ=400 J.
  • 0.45>0.40: claim not feasible: w_(claimed)=450 J>wₘₐₓ=400 J
  • heat rejected q_c=1000-450=550 J. The engine works in a cycle (Δ S_(engine)=0): 4pt] Δ Sᵤₙᵢᵥ=-(qₕ)/(Tₕ)+(q_c)/(T_c)=-(1000)/(500)+(550)/(300)=-2.000+1.833=-0.167 J K⁻¹<0

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: Tₕ=500 K, T_c=300 K, qₕ=1000 J, w=450 J (claimed).

  1. Claimed efficiency: η=(w)/(qₕ)=(450)/(1000)=0.45
  2. Carnot limit: no engine working between two reservoirs can be more efficient than a reversible (Carnot) engine between the same reservoirs (Carnot theorem): ηₘₐₓ=1-(T_c)/(Tₕ)=1-(300)/(500)=0.40, so wₘₐₓ=400 J.
  3. Verdict: 0.45>0.40: claim not feasible: w_(claimed)=450 J>wₘₐₓ=400 J
  4. Entropy check. First law: heat rejected q_c=1000-450=550 J. The engine works in a cycle (Δ S_(engine)=0): 4pt] Δ Sᵤₙᵢᵥ=-(qₕ)/(Tₕ)+(q_c)/(T_c)=-(1000)/(500)+(550)/(300)=-2.000+1.833=-0.167 J K⁻¹<0
  5. Meaning: the cold reservoir must receive at least qₕT_c/Tₕ=600 J to carry away the 2.000 J K⁻¹ taken from the hot one. Rejecting only 550 J would lower the entropy of the universe: the Kelvin–Planck statement is violated (part of the heat would be converted wholly into work).
  6. First law alone is satisfied (1000=450+550); the claim fails only on the second law. A real engine, with friction and finite-rate heat flow, would give even less than 400 J.

Application. the same limit sets the best possible efficiency of power stations: raising the steam temperature or lowering the condenser temperature is the only way to raise ηₘₐₓ; a refrigerator's coefficient of performance is limited in the same way, T_c/(Tₕ-T_c).

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Model answer to Q2(b), 2020 Paper I: solution page 8
Solution page 8 of 45
Read the whole 2020 solution2020 question paper

All parts of Question 2