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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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  1. peterespenshade
    26.10.2017 at 23:33

    I believe your first dS equation is usually referred to as the fundamental thermodynamic relation instead of the 1st law. This relation can be proven using the 1st and 2nd laws.

    • 28.10.2017 at 20:07

      It’s actually a combination of the first and second laws.

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