Question

heat engine perform work at the rate of 500 KW the efficiency of the engine is 30% calculate the loss of heat per hour?
​

Answer 1

Answer:

Answer: 126 \times 10^4 kJ/h126×10

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⇒70%of500kJ/s=

100

70

×500kJ/s×3600s/h=126×10

4

kJ/h

Hence, the loss of heat per hour is 126 \times 10^4 kJ/h126×10

4

kJ/h

Explanation:

hope it helps mark me as brainlist