A steam engine operating between a boiler temperature of 220 ${ }^{\circ} \mathrm{C}$ and a condenser temperature of $35.0^{\circ} \mathrm{C}$ delivers $8.00 \mathrm{hp}$. If its efficiency is $30.0$ percent of that for a Carnot engine operating between these temperature limits, how many calories are absorbed each second by the boiler? How many calories are exhausted to the condenser each second?
$$\text { Actual efficiency }=(0.30) \text { (Carnot efficiency) }=(0.300)\left(1-\frac{308 \mathrm{~K}}{493 \mathrm{~K}}\right)=0.113$$
We can determine the input heat from the relation for the efficiency
Efficiency $=\frac{\text { Output work }}{\text { Input heat }}$
and so every second
$$\text { Input heat } / \mathrm{s}=\frac{\text { Output work } / \mathrm{s}}{\text { Efficiency }}=\frac{(8.00 \mathrm{hp})(746 \mathrm{~W} / \mathrm{hp})\left(\frac{1.00 \mathrm{cal} / \mathrm{s}}{4.184 \mathrm{~W}}\right)}{0.113}=12.7 \mathrm{kcal} / \mathrm{s}$$
To find the energy rejected to the condenser, we use the law of conservation of energy:
Thus,
$$
\begin{aligned}
\text { Input energy } &=(\text { Output work })+(\text { Rejected energy }) \\
\text { Rejected energy } / \mathrm{s} &=(\text { Input energy } / \mathrm{s})-(\text { Output work } / \mathrm{s}) \\
&=(\text { Input energy } / \mathrm{s})-(\text { Input energy } / \mathrm{s})(\text { Efficiency }) \\
&=(\text { Input energy } / \mathrm{s})[1-(\text { Efficiency })] \\
&=(12.7 \mathrm{kcal} / \mathrm{s})(1-0.113)=11.3 \mathrm{kcal} / \mathrm{s}
\end{aligned}
$$