Question

A heat engine having an efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5. The energy absorbed from low temperature reservoir by the refrigerator for each kJ of energy absorbed from high temperature source by the engine is

• A. 0.14 kJ
• B. 0.71 kJ
• C. 3.5 kJ
• D. 7.1 kJ

Answer 1

The correct option is C 3.5 kJ

Given data:
$\eta =70\%|=0.70$

$(COP{)}_R=5$

$Q_1=1kJ$

$Q_3=?$




For heat engine,

$\eta =1-\frac{Q_2}{Q_1}$

$0.70=1-\frac{Q_2}{1}$

or
$Q_2=0.3kJ$

$W=Q_1-Q_2=1-0.3=0.7kJ$

For refrigerator,
$(COP{)}_R=\frac{Q_3}{W}$

$5=\frac{Q_3}{0.7}$

or $Q_3=0.7\times 5=3.5kJ$