Question

A heat engine perform work at the rate of 500kw ,the efficiency of engine is 30%.calculate the loss of heat at per hour

Answer 1

Answer: $126 \times 10^4 kJ/h$

A heat engine perform work at the rate of 500 kW

Rate of doing work is 500 kW = 500 kJ/s

The efficiency of engine is 30%. It means 70% of the energy is lost as heat.

$\Rightarrow 70\% of 500 kJ/s=\frac{70}{100} \times 500 kJ/s \times 3600 s/h=126\times 10^4 kJ/h$

Hence, the loss of heat per hour is $126 \times 10^4 kJ/h$