Question

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and energy is expelled as heat Q2 at a lower temperature T2 = 430 K. The second stage absorbs that energy as heat Q2, does work W2, and expels energy as heat Q3 at a still lower temperature T3 = 240 K. What is the efficiency of the engine?

Answer 1

Efficiency = 52%

Step-by-step explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

`\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}`

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

`\eta=((Q_1-Q_2)\ +\ (Q_2-Q_3))/(Q_1)}`

or

`\eta=1-((Q_1-Q_3))/(Q_1)}`

or

`\eta=1-((Q_3))/(Q_1)}`

also,

`(Q_1)/(T_1)=(Q_2)/(T_2)=(Q_3)/(T_3)`

or

`(T_3)/(T_1)=(Q_3)/(Q_1)`

thus,

`\eta=1-((T_3))/(T_1)}`

thus,

`\eta=1-((240\ K))/(500\ K)}`

or

`\eta=0.52`

or

Efficiency = 52%