# Homework Help: Thermodynamics - entropy

1. Feb 18, 2005

### Kelvin

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:

$$$\begin{gathered} \Delta S = \Delta S_{heat} + \Delta S_{rod} + \Delta S_{cool} \hfill \\ \Delta S_{heat} = \frac{Q} {T} = \frac{{ - 1200{\text{ J}}}} {{130 + 273{\text{ K}}}} \hfill \\ \Delta S_{cool} = \frac{Q} {T} = \frac{{1200{\text{ J}}}} {{24 + 273{\text{ K}}}} \hfill \\ \end{gathered}$$$

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.

2. Feb 18, 2005

### xanthym

The problem begins with the Rod already connected as indicated and already in a dynamically steady-state condition. Therefore, no state change occurs in the Rod for the thermodynamic processes of this problem. Hence:

$$:(1): \ \ \ \ \Delta S = \Delta S_{HOT} + \Delta S_{rod} + \Delta S_{COLD}$$

$$:(2): \ \ \ \ \Delta S_{HOT} = \frac {\Delta Q} {T} = \frac {-1200} {130 + 273}$$

$$:(3): \ \ \ \ \Delta S_{COLD} = \frac {\Delta Q} {T} = \frac {1200} {24 + 273}$$

$$:(4): \ \ \ \ \Delta S_{rod} = 0$$

$$:(5): \ \ \ \ \Delta S = (+1.063 \ J)$$

~~

Last edited: Feb 18, 2005
3. Feb 18, 2005

### Duarh

From the given information, I would assume that the rod doesn't change at all during this process so it can't have any change in entropy. If there is some constant heat flow across the rod, its temperature is ill-defined but you can't really say it's changing in any way, so all the entropy would be due solely to the changes in the heat reservoirs.

I'm fairly new to thermo, though, so a second opinion would be good.

edit: ah, xanthym, great

4. Feb 19, 2005

### Kelvin

thank you xanthym and duarh!

In my mind I have unconsciously assumed that the rod is originally at room temperature before connected to the reserviors, so there should be temperature change.

I see your points ~ thanks a lot :)