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

Answer:

Efficiency = 52%

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,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

or

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

or

$\eta=0.52$

or

Efficiency = 52%