Question

A heat engine of efficiency 40% takes 10 kJ of heat energy every second. It is used to lift a weight of 2000 kg. The height to which the weight can be raised in 30 sec is [Take g = 10 ms²]

Answer 1

Heat engine takes 10 kj heat per second.
efficiency of heat engine = 40%
net heat taken = 10×40/100 = 4 kj

As energy required to lift the water at the height 10 m = mgh = 1000×10×10 = 100000 j =
= 100 kj
4 kj is taken in one second
100 kj will take in 100/4 = 25 seconds.