Question

35. A heat engine receives heat in the amount of Qh= 775 kJ from a high thermal reservoir and delivers Wnet= 210 kJ of work per cycle.

1. Write an expression for the efficiency of the engine.
2. What is the efficiency?
3. Write an expression for the amount of energy required to be rejected into the low temperature reservoir.
4. How much energy, in kilojoules, is rejected into the low temperature reservoir?

Answer 1

Given,

The amount of heat received by the heat engine is, Qh = 775 KJ

The work, Wnet = 210 KJ

a)

The expression for the efficiency of the engine is, = Wnet / Qh

b)

The efficiency,   = Wnet / Qh

= 210 / 775

= 0.271

c)

The expression for the amount of energy required to be rejected into low temperature reservoir is,

Qe = Qh - Wnet

d)

The amount of energy required to be rejected into a low temperature reservoir is,

Qe = Qh - Wnet

Qe = 775 - 210

Qe = 565 KJ