- #1
Kelvin
- 52
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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.
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
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.
![Cry :cry: :cry:](/styles/physicsforums/xenforo/smilies/cry.png)
Thanks in advance