Thermodynamics

JEE Main level, single correct, medium

188 of 190

Question 188 of 190

A solid body of constant heat capacity 1J(℃)−11J{\left(℃\right)}^{-1} 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 the same amount of heat. (ii) Sequentially keeping in contact with 88 reservoirs such that each reservoir supplies the same amount of heat. In both, cases the body is brought from initial temperature 100K100K to final temperature 200K200K . Entropy change of the body in the two cases respectively is: Note: This question was awarded as a bonus since temperatures were given in centigrade instead of in Kelvin. Proper corrections are made in the question to avoid it.

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Answer: (C) ln⁡2,ln⁡2\ln 2,\ln 2

Change in entropy,
dS=dQTΔQ=heatsupplied=CΔTC=J(∘C)−1dQ=CdtdS=CdTTIntergratingbothsides∫SiSfdS=C∫dTT⇒Sf−Si=ΔS=C.ln⁡T∣100200⇒ΔS=C[ln⁡200−ln⁡100]=Cln⁡2andC=J(∘C)−1⇒ΔS=ln⁡2Entrpoy change in same for both cases as C is constant, andtemperature change (i.e. from 100 to 200) is same.\begin{matrix}dS=\dfrac{dQ}{T} \\ \\ \Delta Q= \text{heat} \text{supplied}=C\Delta T {C=J{\left(^{\circ}C\right)}^{-1}} \\ \\ dQ=Cdt \\ \\ dS=\dfrac{CdT}{T} \\ \\ \text{Intergrating} \text{both} \text{sides} \\ \\ \int _{{S}_{i}}^{{S}_{f}}dS=C\int \dfrac{dT}{T} \\ \\ \Rightarrow {S}_{f}-{S}_{i}=\Delta S=C.\ln T{|}_{100}^{200} \\ \\ \Rightarrow \Delta S=C[\ln 200-\ln 100]=C\ln 2 \\ \\ \text{and} C=J{\left(^{\circ}C\right)}^{-1} \\ \\ \Rightarrow \Delta S=\ln 2 \\ \\ \text{Entrpoy change in same for both cases as C is constant, and} \\ \text{temperature change (i}\text{.e}\text{. from 100 to 200) is same}\text{.}\end{matrix}

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