Answer
$420^{\circ} C$.
Work Step by Step
The maximum efficiency (Carnot efficiency) is given by equation 15–5. Make sure to measure temperatures in kelvins. The efficiency is also $W/Q_H$. Equate these two expressions and solve for the exhaust temperature.
$$e_{ideal}=1-\frac{T_L}{T_H}=\frac{W}{Q_H}$$
$$T_L=T_H(1-\frac{W}{Q_H})= T_H(1-\frac{W/t}{Q_H/t}) $$
$$ = (520+273K)(1-\frac{5.2\times10^5\;J/s}{(950kcal/s)(4186J/kcal)})=689K\approx 420^{\circ} C $$
