Differentials of Entropy for Air and Water at Different Temperatures

[yamata1]

Homework Statement

A bottle of water is removed from a cupboard at the initial temperature $T_i$. In the ambient air of constant temperature $T_0$, it warms or cools to reach equilibrium at the final temperature
$T_f = T_0$. The heat capacity of the plastic of the bottle is neglected compared to that of the water.
The thermal expansion of the water is also neglected.


a)The heat capacity C of the water depends on the temperature. Express entropy variations
$\Delta S$ of water, $\Delta S_{th}$ of the air and $\Delta S_{univ}$ of the universe in the form of integrals.

b) Show that, whatever the temperatures, $\Delta S_{univ}\geq 0$.

Relevant Equations

$dS=\frac{dQ}{T}$

for a)$\Delta S=\mp \int_{T_i}^{T_0}\frac{C(T)}{T}dT$ and $\Delta S_{th}=\int_{T_i}^{T_0}\frac{dQ}{T_0}dT$ so $S_{univ}=\Delta S_{th}+\Delta S$.

What is $dQ$ equal to ? I don't know how to answer question b).

Thank you for your help.

 

[haruspex]

Your integral for $\Delta S_{th}$ makes no sense. You cannot have two 'd' terms in a single integral.
Hint: you do not need that integral at all.

 

[yamata1]

Your integral for $\Delta S_{th}$ makes no sense. You cannot have two 'd' terms in a single integral.
Hint: you do not need that integral at all.

Indeed,Is $\Delta S_{th}=\mp\frac{Q}{T_0}=\mp\int_{T_i}^{T_0}\frac{C(T)}{T_0}dT$ correct ?
If so then question b) is $\Delta S_{univ}=\int_{T_i}^{T_0}C(T)(\frac{1}{T}-\frac{1}{T_0})dT$ when the water heats up and $\Delta S_{univ}=\int_{T_i}^{T_0}C(T)(\frac{1}{T_0}-\frac{1}{T})dT$ when it cools off .Both of these integrals are positive since C(T) is strictly positive.

 

[kuruman]

Hint: you do not need that integral at all.

True, but the question (part a) asks to express the entropy changes "in the form of integrals."

 

[haruspex]

Indeed,Is $\Delta S_{th}=\mp\frac{Q}{T_0}=\mp\int_{T_i}^{T_0}\frac{C(T)}{T_0}dT$ correct ?
If so then question b) is $\Delta S_{univ}=\int_{T_i}^{T_0}C(T)(\frac{1}{T}-\frac{1}{T_0})dT$ when the water heats up and $\Delta S_{univ}=\int_{T_i}^{T_0}C(T)(\frac{1}{T_0}-\frac{1}{T})dT$ when it cools off .Both of these integrals are positive since C(T) is strictly positive.

Since the ambient temperature does not change you cannot express the air's entropy change as an integral dT. What is the other option?

 

[haruspex]

True, but the question (part a) asks to express the entropy changes "in the form of integrals."

Ah, quite so.

 

[rude man]

Compare differentials dS for the air and for the water thruout the range of T.