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JEE Main 2015 · 4 Apr · Shift 1, Physics question 10

A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways :

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat.

In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is :

Choose one option
Answer and worked solution

Official answer: Bonus: full marks to everyone

Entropy is a state function. In both cases the body goes from 373 K to 473 K, so
ΔS=Cln⁡473373≈0.24 J/K\Delta S=C\ln\frac{473}{373}\approx0.24\text{ J/K}
in each case, however many reservoirs are used. No option gives this (the options take ln⁡2\ln2, as if the temperatures were 100 K and 200 K), and the question was given marks to all.

Topics:Second law and entropy
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