## 4F.3 Reversible Process

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

KDang_1D
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### 4F.3 Reversible Process

Calculate the change in entropy of a block of copper at 25ºC that absorbs 65J of energy from a heater.

Why can we assume that this is a reversible process?

Nick Fiorentino 1E
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Joined: Wed Sep 18, 2019 12:16 am

### Re: 4F.3 Reversible Process

Maybe it's because the heat can be given off from the copper block?

Brooke Yasuda 2J
Posts: 102
Joined: Sat Jul 20, 2019 12:17 am

### Re: 4F.3 Reversible Process

We can approximate here and assume that it is a reversible reaction because the surroundings are so large that any change in heat will have little to no effect on the the surroundings.