Question 20.4: Analyzing a Carnot refrigerator If the cycle described in Ex......

IDENTIFY and SET UP This problem uses the ideas of Section 20.3 (for refrigerators in general) and the above discussion of Carnot refrigerators. Equation (20.9)

K=\frac{\left|Q_{\mathrm{C}}\right|}{|W|}=\frac{\left|Q_{\mathrm{C}}\right|}{\left|Q_{\mathrm{H}}\right|-\left|Q_{\mathrm{C}}\right|}     (20.9)

gives the coefficient of performance K of any refrigerator in terms of the heat Q_{\mathrm{C}} extracted from the cold reservoir per cycle and the work W that must be done per cycle.

EXECUTE In Example 20.3 we found that in one cycle the Carnot engine rejects heat Q_{\mathrm{C}}=-346 \mathrm{~J} to the cold reservoir and does work W=230 \mathrm{~J}. When run in reverse as a refrigerator, the system extracts heat Q_{\mathrm{C}}=+346 \mathrm{~J} from the cold reservoir while requiring a work input of W=-230 J. From Eq. (20.9),

\begin{aligned} K & =\frac{\left|Q_{\mathrm{C}}\right|}{|W|}=\frac{346 \mathrm{~J}}{230 \mathrm{~J}} \\ & =1.50 \end{aligned}

Because this is a Carnot cycle, we can also use Eq. (20.15):

\begin{aligned} K & =\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}=\frac{300 \mathrm{~K}}{500 \mathrm{~K}-300 \mathrm{~K}} \\ & =1.50 \end{aligned}

EVALUATE Equations (20.14)

e_{\text {Carnot }}=1-\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}}=\frac{T_{\mathrm{H}}-T_{\mathrm{C}}}{T_{\mathrm{H}}}       (20.14)

and (20.15) show that e and K for a Carnot cycle depend only on T_{\mathrm{H}} and T_{\mathrm{C}}, and we don’t need to calculate Q and
W. For cycles containing irreversible processes, however, these two equations are not valid, and more detailed calculations are necessary.

KEYCONCEPT A refrigerator takes heat in from a cold reservoir and rejects heat to a hot reservoir. Work must be done on the refrigerator to make this happen. The coefficient of performance K of the refrigerator equals the amount of heat that is rejected to the hot reservoir divided by the amount of work that must be done on the refrigerator. The greater the value of K, the more heat can be removed from the cold reservoir for a given expenditure of work.