Calculating Entropy Change of a Wire Conduction

[pizzafihop]

Homework Statement

Each end of a metal wire is in thermal contact with a different heat reservoir. Reservoir 1 is at a temperature of 752 K, and reservoir 2 is at a temperature of 345 K. Compute the total change in entropy that occurs from the condustion of 1096 J of heat through the wire.

Homework Equations

dS = dQ / T

The Attempt at a Solution

Which T should I use? Should I average them? Use only one? Divide one by the other?

 

[Andrew Mason]

Homework Statement

Each end of a metal wire is in thermal contact with a different heat reservoir. Reservoir 1 is at a temperature of 752 K, and reservoir 2 is at a temperature of 345 K. Compute the total change in entropy that occurs from the condustion of 1096 J of heat through the wire.

Homework Equations

dS = dQ / T

The Attempt at a Solution

Which T should I use? Should I average them? Use only one? Divide one by the other?

First of all, you have to ignore the entropy change of the wire. Then it is just a matter of heat flowing out of the hot reservoir and heat flowing into the cold reservoir without changing the temperature of either. Calculate the entropy change of the hot reservoir. Calculate the entropy change of the cold reservoir. Add them together (be careful of the signs - flow into is positive/flow out is negative).

AM

 

[pizzafihop]

Thanks, I didn't realize I had to split the formula in 2 for each reservoir.

 

[Andrew Mason]

Thanks, I didn't realize I had to split the formula in 2 for each reservoir.

To calculate the total entropy change you have to sum the entropy changes of the parts.

Generally, you would divide the entire system into infinitessimal slices and do an integral of the heat flows/surface temperature into and out of each slice. In this case there is no temperature gradient within the reservoir so you can look at the reservoir as a whole. All the heat leaves or enters the reservoir at the same temperature. So just determine the entropy change for each reservoir and add them together to get the total entropy change for the system.

Once the wire heats up there is a stable temperature gradient along the wire so its thermodynamic state does not change. You are then left with only the entropy changes in each reservoir.

AM