Question

The efficiency of heat engine is 10.if heat energy consumed by the engine is 250KJ, how high would this engine lift an object of mass 1 kg from the ground?

Answer 1

Given, H = 250KJ = 250000 J, n = 10, 

We know that n = W/H * 100

                       W = 10 * 250000/100

                             =   25000


We know that W=mgh

                 25000=1 * 10 * h
 
                    h = 2500 m.


Hope this helps!