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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?

Respuesta :

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:

[tex]\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}[/tex]

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

[tex]\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}[/tex]

or

[tex]\eta=1-\frac{(Q_1-Q_3)}{Q_1}}[/tex]

or

[tex]\eta=1-\frac{(Q_3)}{Q_1}}[/tex]

also,

[tex]\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}[/tex]

or

[tex]\frac{T_3}{T_1}=\frac{Q_3}{Q_1}[/tex]

thus,

[tex]\eta=1-\frac{(T_3)}{T_1}}[/tex]

thus,

[tex]\eta=1-\frac{(240\ K)}{500\ K}}[/tex]

or

[tex]\eta=0.52[/tex]

or

Efficiency = 52%

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