# Chapter 19 - Exercises and Problems - Page 363: 63

61%; this is much larger than the 25 percent actual efficiency.

#### Work Step by Step

As we know that $T_C=90F^{\circ}=(90-32)\frac{5}{9}+273K=305K$ $e_{Carnot}=(1-\frac{T_C}{T_K})\times 100\%$ We plug in the known values to obtain: $e_{Carnot}=(1-\frac{305K}{783K})\times 100\%$ $e_{Carnot}=61\%$ It is clear that the efficiency is greater than actual efficiency that is $25\%$.

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