Entropy change for water in contact with a reservoir

[Parzeevahl]

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?

 

[mjc123]

Note that it says dS, not ΔS - this is an infinitesimal change. As the temperature of the water is changing, to get ΔS you have to integrate dS over the temperature range.
Your calculation gives the entropy change of the reservoir (with the wrong sign) because the temperature of the reservoir doesn't change, so ΔS_res = Q_res/T_res, and Q_res = -Q_water.

 

[Abhishek11235]

Hint: What exactly is heat reservoir?
For 2nd question ,try bringing water from the given state quasi-statically to final state.

 

[BvU]

Hello Parzeevahl, $\qquad$ $\qquad$ !

Look up the specific entropy of water at 273 and at 373 and compare with your result.
What is the crucial difference in this process when you compare the container with the reservoir ?

 

[Chestermiller]

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)

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)

These equations are for an ideal gas, not for liquid water. And, even for that case, they are missing a factor of n in front of the heat capacities.

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.

This result was not obtained by integrating the equation. It was obtained by dividing the heat transferred by the reservoir temperature. That gives the entropy change of the reservoir, if corrected in sign, as pointed out by mjc123.

What do you get if you calculate the entropy change of the water by integrating the equation that you gave?

For a cookbook recipe on how to determine, in general, the entropy change of a system that experiences an irreversible process like this (including worked examples), please see my Physics Forums Insights article at https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/