Change in Temperature for Stretched Surface

[GL_Black_Hole]

Homework Statement

The surface tension of a layer of water obeys $\sigma = a- bT$, where $T$ is the temperature. Find the change in temperaure, $\Delta T$ when the area is increased isentropically.

Homework Equations

$dU = dQ -dW$ , $dW = \sigma dA$, $dU = C_A dT + [\sigma -T(\frac{\partial\sigma}{\partial T})] dA$

The Attempt at a Solution

Evaluating $dU$ gives $dU = C_A dT + [\sigma -T(-b)] dA$ which means $dU = C_A dT + adA$, so rearranging the first law of thermodynamics for $dS$ I get:
$dS = \frac{dU + \sigma dA}{T} = \frac{C_A}{T} dT + \frac{a}{T} dA + \frac{a}{T}dA - bdA$.

I know that integrating dS has to give zero but don't see how to extract $\Delta T$ from this.

 

[Chestermiller]

Homework Statement

The surface tension of a layer of water obeys $\sigma = a- bT$, where $T$ is the temperature. Find the change in temperaure, $\Delta T$ when the area is increased isentropically.

Homework Equations

$dU = dQ -dW$ , $dW = \sigma dA$, $dU = C_A dT + [\sigma -T(\frac{\partial\sigma}{\partial T})] dA$

The Attempt at a Solution

Evaluating $dU$ gives $dU = C_A dT + [\sigma -T(-b)] dA$ which means $dU = C_A dT + adA$, so rearranging the first law of thermodynamics for $dS$ I get:
$dS = \frac{dU + \sigma dA}{T} = \frac{C_A}{T} dT + \frac{a}{T} dA + \frac{a}{T}dA - bdA$.

I know that integrating dS has to give zero but don't see how to extract $\Delta T$ from this.

You have $dU=TdS+\sigma dA=(a-bT)dA=C_AdT+adA$