The temperature of a Carnot refrigerator’s cold reservoir is negative 73.0 degrees Celsius. The hot reservoir has a temperature of 270 degrees Celsius. What is the coefficient of performance of the Carnot refrigerator?
When we talk about a Carnot refrigerator or a Carnot engine, that’s the name that refers to the most efficient heat cycle that’s allowed by physical law. In this case, we want to solve for the coefficient of performance of this Carnot refrigerator. We can abbreviate this COP.
One way to describe the coefficient of performance is to say that it’s equal to the cooling effect divided by the work input into this engine. The more we’re able to cool an interior space down for a given amount of work, the higher our coefficient of performance.
We can also say that our COP is equal to the heat absorbed by our system divided by the heat rejected minus that heat absorbed. In the problem statement, we’re told about a cold reservoir as well as a hot reservoir. The cold reservoir is the one doing the heat absorbing. And the hot reservoir is the one doing the heat rejecting.
As a third way of expressing COP then, we can say it’s equal to the temperature of the cold reservoir divided by the temperature of the hot minus the temperature of the cold reservoir. We’re given those temperatures in the problem statement. And to solve for the COP, we’ll use those temperatures, converted to the Kelvin scale, in this relationship.
To do this conversion, we’ll take our temperatures in degrees Celsius and add 273.15 to it to create a temperature in Kelvin. We do this for both, the cold and the hot temperature. And when we calculate this fraction, we find it’s equal to 0.58. That’s the coefficient of performance of this Carnot refrigerator.