An ideal Carnot refrigerator with a performance coefficient (COP)of 5.0 cools items inside of it to $5.0 ^ { \circ } \mathrm { C }$ What is the high temperature needed to operate this refrigerator?

Question

Quizplus · Accepted Answer

The COP of a Carnot refrigerator is given by $\text{COP} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}}$, where $T_{\text{cold}}$ and $T_{\text{hot}}$ are the temperatures of the cold and hot reservoirs, respectively, in Kelvin. Rearranging for $T_{\text{hot}}$, we get $T_{\text{hot}} = T_{\text{cold}} + \frac{T_{\text{cold}}}{\text{COP}}$. Converting $5.0 ^{\circ}C$ to Kelvin gives $278.15 K$. Substituting $T_{\text{cold}} = 278.15 K$ and $\text{COP} = 5.0$ into the equation gives $T_{\text{hot}} = 278.15 K + \frac{278.15 K}{5} = 278.15 K + 55.63 K = 333.78 K$, which is approximately $60.63 ^{\circ}C$, closest to $61 ^{\circ}C$.

An Ideal Carnot Refrigerator with a Performance Coefficient (COP)of 5 5.0∘C5.0 ^ { \circ } \mathrm { C }5.0∘C

An Ideal Carnot Refrigerator with a Performance Coefficient (COP)of 5 $5.0 ^ { \circ } \mathrm { C }$