A brass rod is in thermal contact with a heat reservoir at 130 degree C at one end and a heat reservoir at 24.0 degree C at the other end. Compute the total change in the entropy arising from the process of conduction of 1200 J of heat through the rod.(adsbygoogle = window.adsbygoogle || []).push({});

My attempt is:

[tex]

\[

\begin{gathered}

\Delta S = \Delta S_{heat} + \Delta S_{rod} + \Delta S_{cool} \hfill \\

\Delta S_{heat} = \frac{Q}

{T} = \frac{{ - 1200{\text{ J}}}}

{{130 + 273{\text{ K}}}} \hfill \\

\Delta S_{cool} = \frac{Q}

{T} = \frac{{1200{\text{ J}}}}

{{24 + 273{\text{ K}}}} \hfill \\

\end{gathered}

\]

[/tex]

so far, am I correct? If yes, how can I calculate the entropy change of the rod?

I know it is a irreversible process. but without knowing the intial states (p, V, T), it is impossible to construct a reversible one connecting the intial and final states and then make use of the fact that entropy is a state function.

so I think there must be an indirect method to know the entropy change of the rod.

Please give me some hints coz I have read the entropy chapter of my textbook for many times and I'm still not sure how to approach it.

Thanks in advance

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# Homework Help: Thermodynamics - entropy

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