- #1

Parzeevahl

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- Homework Statement
- 1 kg of water at 273 K is brought into contact with a heat reservoir at 373 K. When the water has reached 373 K, what is the entropy change of the water, of the heat reservoir, and of the universe?

- Relevant Equations
- dS=Cp*(dT/T)-nR*(dP/P)

dS=Cv*(dT/T)+nR*(dV/V)

**Problem Statement:**1 kg of water at 273 K is brought into contact with a heat reservoir at 373 K. When the water has reached 373 K, what is the entropy change of the water, of the heat reservoir, and of the universe?

**Relevant Equations:**dS=Cp*(dT/T)-nR*(dP/P)

dS=Cv*(dT/T)+nR*(dV/V)

I am assuming that the question means the water is in a container or something, so its pressure and/or volume shouldn't change on heating. Then dP=dV=0.

Now, at 273 K and 373 K, the specific heat of water, Cp, is nearly the same, So, I take it as a constant = 4217 J/(kg.K).

So, dS=Cp*(dT/T)

I have dT=373K-273K=100K, T=373K.

Putting values, I get dS=1130 J/K.

But I'm confused whether this is the entropy change of the

*reservoir*or the

*water*. The answer given in the book says that it is the entropy change of the

*reservoir*, but I can't understand why. Also, if this

*is*the entropy change of the reservoir, when how do I calculate that of the water?