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Question

(A) We know that the change in entropy is given by $\Delta S = \frac{q_{rev}}{T}$.Given that the quantity of heat $q_{rev}$ is the same in both cases,

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$\Delta S_{1} < \Delta S_{2}$

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(A) We know that the change in entropy is given by $\Delta S = \frac{q_{rev}}{T}$.Given that the quantity of heat $q_{rev}$ is the same in both cases,we have $\Delta S_{1} = \frac{q_{rev}}{T_{1}}$ and $\Delta S_{2} = \frac{q_{rev}}{T_{2}}$.Since $T_{1} > T_{2}$,the denominator in the expression for $\Delta S_{1}$ is larger than the denominator in the expression for $\Delta S_{2}$.Therefore,$\Delta S_{1} < \Delta S_{2}$.

When the same quantity of heat is absorbed by a system at two different temperatures $T_{1}$ and $T_{2}$,such that $T_{1} > T_{2}$,the changes in entropy are $\Delta S_{1}$ and $\Delta S_{2}$ respectively. Then:

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