Question

A Carnot cooler operates with COP = 11, whose ambient temperature is 300K. Determine the temperature at which the refrigerator absorbs heat.

Answer 1

275 Kelvin

Step-by-step explanation:

Coefficient of Performance=11

`T_H=\text {Absolute Temperature of high temperature reservoir=300 K}`

`T_L=\text {Absolute Temperature of low temperature reservoir}`

`\text {Coefficient of performance for carnot cooler}\\=\frac {T_L}{T_H-T_L}\\\Rightarrow 11=(T_L)/(300-T_L)\\\Rightarrow 11(300-T_L)=T_L\\\Rightarrow 3300-11T_L=T_L\\\Rightarrow 3300=T_L+11T_L\\\Rightarrow 3300=12T_L\\\Rightarrow T_L=\frac {3300}{12}\\\Rightarrow T_L=275\ K\\\Therefore \text{Temperature at which the refrigerator absorbs heat=275 Kelvin}`