Essential University Physics: Volume 1 (3rd Edition)

a) $T_h = \frac{T_{h0}}{e^({\frac{P_0t}{-mcT_c+mcT_{h0}}})}$ b)$t=\frac{(-mcT_c+mcT_{h0})ln(\frac{T_{h0}}{T_c})}{P_0}$
a) We know the following equations: $P=P_0\frac{T_h-T_c}{T_{h0}-T_c}$ $dW = Pdt$ $Pdt = \frac{(-mc\space dT_h)(T_h-T_c)}{T_h}$ Combining these two equations and taking the integral, we simplify to find that the value of $T_h$ is: $T_h = \frac{T_{h0}}{e^({\frac{P_0t}{-mcT_c+mcT_{h0}}})}$ b) First, we simplify the above expression for t: $T_h = \frac{T_{h0}}{e^({\frac{P_0t}{-mcT_c+mcT_{h0}}})}$ $T_h(e^{\frac{P_0t}{-mcT_c+mcT_{h0}}}) = T_{h0}$ $(e^{\frac{P_0t}{-mcT_c+mcT_{h0}}}) = \frac{T_{h0}}{T_h}$ $\frac{P_0t}{-mcT_c+mcT_{h0}}=ln(\frac{T_{h0}}{T_h})$ $P_0t=(-mcT_c+mcT_{h0})ln(\frac{T_{h0}}{T_h})$ $t=\frac{(-mcT_c+mcT_{h0})ln(\frac{T_{h0}}{T_h})}{P_0}$ We know that the engine will not longer work when $T_h$ and $T_c$ are equal, so it follows: $t=\frac{(-mcT_c+mcT_{h0})ln(\frac{T_{h0}}{T_c})}{P_0}$