The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is

Question

The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4 . It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine is being fully used to run a heat pump which has a coefficient of performance 10 . The hot reservoir of the heat pump is at a temperature of 300 K . Which of the following statements is/are correct?\

Professor PC Thomas Classes · Answer

(a, b, c) : $T_{H}=1000 \mathrm{~K}$,\ $\eta=0.4, Q_{H}=150 \mathrm{~J}$\ $C O P=10$, heat pump, $T_{H}{ }^{\prime}=300 \mathrm{~K}$\ For engine,\ $\eta=1-\frac{T_{C}}{T_{H}} \Rightarrow 0.4=1-\frac{T_{C}}{1000}$\ $\Rightarrow \quad T_{C}=600 \mathrm{~K}$\ Now, $\frac{Q_{H}}{Q_{C}}=\frac{T_{H}}{T_{C}}$\ $$ \begin{aligned} \Rightarrow \frac{150}{Q_{C}} & =\frac{1000}{600} \Rightarrow Q_{C}=90 \mathrm{~J} \ W & =Q_{H}-Q_{C}=150-90=60 \mathrm{~J} \end{aligned} $$\ So, COP $=\frac{	ext { Output }}{	ext { Input }} \Rightarrow 10=\frac{Q_{3}}{W}$ $$ Q_{3}=10 	imes 60=600 \mathrm{~J} $$ So, heat supplied to hot reservoir of heat pump, $$ \begin{aligned} & Q_{4}=W+600=600+60=660 \mathrm{~J} \ & \eta=\frac{1}{\mathrm{COP}}=0.1 \ & \left(1-\frac{T_{3}}{300}ight)=0.1 \Rightarrow T_{3}=270 \mathrm{~K} \end{aligned} $$