## Essential University Physics: Volume 1 (3rd Edition)

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 $25\%$.