Question

A carnot engine having an efficiency of $\eta =\frac{1}{5}$ is used as a heat engine. Now it is used as a refrigerator. If work done by compressor is $40\text{J}$, the amount of energy delivered to surroundings is

• A. $200\text{J}$
• B. $160\text{J}$
• C. $120\text{J}$
• D. $50\text{J}$

Answer 1

The correct option is A $200\text{J}$

$\underset{\text{(heat engine)}}{\eta }=1-\frac{T_L}{T_H}=\frac{1}{5}$
$\Rightarrow \frac{T_L}{T_H}=\frac{4}{5}\text{if}{T}_H=X,T_L=\frac{4X}{5}$


$\underset{\text{(refrigerator)}}{\text{C.O.P}}=\frac{T_L}{T_H-T_L}=\frac{\frac{4X}{5}}{X-\frac{4X}{5}}=\frac{\frac{4X}{\text{/}5}}{\frac{X}{\text{/}5}}=4$
We know that $\text{C.O.P}=\frac{Q_{absorbed}}{W}$
$\Rightarrow 4=\frac{Q_{absorbed}}{40}$
$Q_{absorbed}=4\left(40\right)=160\text{J}$
We know that the heat delivered to the surroundings is
$Q_{delivered}=Q_{absorbed}+W$
$Q_{delivered}=40+160=200\text{J}$