Physics 4311: Thermal Physics – Test 2 Solution

Problem 1: TD potentials and Maxwell relations (42 points)

a)

$H=U-\mathit{fL},\mathit{dH}=\mathit{TdS}-\mathit{Ldf}$
$F=U-\mathit{TS},\mathit{dF}=-\mathit{SdT}+\mathit{fdL}$
$G=U-\mathit{TS}-\mathit{fL},\mathit{dG}=-\mathit{SdT}-\mathit{Ldf}$

b)

from $U$: ${\left(\frac{\mathit{\partial f}}{\mathit{\partial S}}\right)}_L={\left(\frac{\mathit{\partial T}}{\mathit{\partial L}}\right)}_S$
from $H$: ${\left(\frac{\mathit{\partial L}}{\mathit{\partial S}}\right)}_f=-{\left(\frac{\mathit{\partial T}}{\mathit{\partial f}}\right)}_S$
from $F$: ${\left(\frac{\mathit{\partial S}}{\mathit{\partial L}}\right)}_T=-{\left(\frac{\mathit{\partial f}}{\mathit{\partial T}}\right)}_L$
from $G$: ${\left(\frac{\mathit{\partial S}}{\mathit{\partial f}}\right)}_T={\left(\frac{\mathit{\partial L}}{\mathit{\partial T}}\right)}_f$

Problem 2: Stretching a rubber band (60 points)

a)

$\mathit{\delta W}=\mathit{fdL}$
$\mathrm{\Delta }W={\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}_{L_0}^{2L_0}\mathit{fdL}={\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}_{L_0}^{2L_0}\mathit{\alpha LTdL}=\frac{3}{2}\mathit{\alpha T}{L}_0^2$

b)

$\mathit{dU}=\mathit{\delta Q}+\mathit{\delta W}$, isothermal $\mathit{dU}=0$, $\mathrm{\Delta }Q=-\mathrm{\Delta }W=-\frac{3}{2}\mathit{\alpha T}{L}_0^2$

c)

$\mathit{dS}=\mathit{\delta Q}∕T$, $\mathrm{\Delta }S=\mathrm{\Delta }Q∕T=-\frac{3}{2}\alpha L_0^2$

Problem 3: Ideal refrigerator (48 points)

The heat egine, running backwards, must remove $Q_{\mathit{loss}}+Q_B$ from the low tempearture reservoir: $Q_l=Q_{\mathit{loss}}+Q_B$

Use $(Q_h+Q_l)∕Q_h=1-T_l∕T_h$ $\Rightarrow$ $Q_l∕Q_h=-T_l∕T_h$ $\Rightarrow$ $Q_h=-(T_h∕T_l)Q_l$

$W=-(Q_h+Q_l)=\left(\frac{T_h}{T_l}-1\right)Q_l=\left(\frac{T_h}{T_l}-1\right)(Q_{\mathit{loss}}+Q_B)$

$W=A\frac{{(T_h-T_l)}^2}{T_l}+Q_B\frac{T_h-T_l}{T_l}$