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## GR0877 Problem 33

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) According to the first law of thermodynamics $dS = \frac{1}{T}dU + \frac{P}{T} dV = mc \frac{dT}{T} + \nu R \frac{dV}{V}$, where $c$ is the specific heat (per one kilogram). Assuming water is incompressible fluid one has $dS = mc \frac{dT}{T}$. Integrating this from $T_1$ to $T_2$ one obtain $mc \ln{\frac{T_2}{T_1}}$.

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