JEE Mains · Physics · STD 11 - 11. thermodynamics
Water of mass \(m\) gram is slowly heated to increase the temperature from \(T_1\) to \(T_2\) The change in entropy of the water, given specific heat of water is \(1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\), is :
- A \(\mathrm{m} \ln \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)\)
- B zero
- C \(\mathrm{m} \ln \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)\)
- D \(\mathrm{m}\left(\mathrm{T}_2-\mathrm{T}_1\right)\)
Answer & Solution
Correct Answer
(A) \(\mathrm{m} \ln \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & d \mathrm{Q}=\mathrm{msdT} \\ & \mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}=\frac{\mathrm{msdT}}{\mathrm{T}} \\ & \Delta \mathrm{S}=\int \frac{\mathrm{msdT}}{\mathrm{T}}=\mathrm{ms} \ln \frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}} \\ & \Delta…
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