A sample of cooler water is placed in a large heat reservoir, mass $m$ and specific heat $c_p$ . I want to compute the change in entropy for the sample and the reservoir ( the environment ). I think I can do this with $$\Delta S_{sample} =\int_{T_H}^{T_C}dS=\int_{T_H}^{T_C}\frac{Q}{T}=\int_{T_H}^{T_C}\frac{c_p m dT}{T}$$

For the reservoir the temperature is constant and its entropy is $$\Delta S_{res}=\frac{Q_{res}}{T_{res}}$$

Where $Q_{res}$ is the heat transferred from the reservoir to the sample. Is the use of calculus above correct?


1 Answer 1


This is an irreversible process, so, to get the change in entropy of the water, you need to devise an alternative reversible process that takes the water between the same initial and final states. Such a process could involve a continuous sequence of constant temperature reservoirs that run rom $T_C$ to $T_H$. The water is gradually contacted with these reservoirs in sequence such that its temperature changes very gradually (a reversible process). In that context, the very right side of the equation you wrote for the water is correct, except for the limits of integration, which, for heating the water, are switched.

For the reservoir at temperature $T_H$, which transfers heat to the water sample, your equation is correct except for the sign which is negative. Plus, $Q_{res}=mC_p(T_H-T_C)$. So the equation should read $$\Delta S_{res}=-\frac{mC_p(T_H-T_C)}{T_H}$$


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