Question

An inventor has built an engine X and claims that its efficiency X is greater than the efficiency of an ideal engine operating between the same two temperatures. Suppose you couple engine X to an ideal refrigerator (Fig. 20-34a) and adjust the cycle of engine X so that the work per cycle it provides equals the work per cycle required by the ideal refrigerator. Treat this combination as a single unit and show that if the inventor’s claim were true$\left(\mathrm{if}{\epsilon }_x>\epsilon \right)$, the combined unit would act as a perfect refrigerator (Fig. 20-34b), transferring energy as heat from the low-temperature reservoir to the high-temperature reservoir without the need for work.

Short answer

The combined unit in the figure $20-34\mathrm{b}$violates the second law of thermodynamics, so the combined unit can't act as a perfect refrigerator without the need for work.

Step-by-step solution

The given data

The efficiency ${\mathrm{\epsilon }}_{\mathrm{x}}$of engine X is ${\mathrm{\epsilon }}_{\mathrm{x}}>\mathrm{\epsilon }$.

Understanding the concept of the second law of thermodynamics

By using equations$\mathbf{20}\mathbf{-}\mathbf{5}\mathbf{,}\mathbf{}\mathbf{20}\mathbf{-}\mathbf{6}$and $\mathbf{20}\mathbf{-}\mathbf{7}\mathbf{,}\mathbf{}\left(\mathrm{if}{\epsilon }_x>\epsilon \right)$we can find if the combined unit would act as a perfect refrigerator, transforming energy as heat from the low-temperature reservoir to the high-temperature reservoir, without the need for work.

Formulae:

The net entropy change per cycle, $∆\mathrm{S}=\frac{\left|{\mathrm{Q}}_{\mathrm{H}}\right|}{{\mathrm{T}}_{\mathrm{H}}}-\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{{\mathrm{T}}_{\mathrm{L}}}$ (1)

where, ${\mathrm{Q}}_{\mathrm{H}}$and ${\mathrm{Q}}_{\mathrm{L}}$are the heats at high and low temperature respectively. Also ${\mathrm{T}}_{\mathrm{H}}$and ${\mathrm{T}}_{\mathrm{L}}$are the high and low temperatures respectively.

The second law of thermodynamics equation, $∆S\ge 0$ (2)

The first law of thermodynamics for a Carnot cycle, $\mathrm{W}=\left|{\mathrm{Q}}_{\mathrm{H}}\right|-\left|{\mathrm{Q}}_{\mathrm{L}}\right|$ (3)

The efficiency of an ideal Carnot cycle, $\mathrm{\epsilon }=\frac{\mathrm{W}}{\mathrm{Q}}$ (4)

Calculation to check the claim of the inventor that is (if εx>ε)

The inventor’s claim implies that less heat (typically from burning fuel) is needed to operate his engine than, say, a Carnot engine for the same magnitude of net work done. Then, from the Figure$20-34\left(\mathrm{a}\right)$using,${\mathrm{\epsilon }}_{\mathrm{x}}>\mathrm{\epsilon }$ we can see that

$\mathrm{Q}{\text{'}}_{\mathrm{H}}<{\mathrm{Q}}_{\mathrm{H}}$

This implies that the ideal refrigerator unit is delivering more heat to the high-temperature reservoir than engine X draws from it. Therefore, using the concept of energy conservation, we can say that this violates the second law of thermodynamics that is the entropy of the system no more remains positive or zero. So, the combined unit can't act as a perfect refrigerator without the need for work.