Question

5- The inside of a Carnot refrigerator is maintained at a temperature of 277 K, while the temperature in the kitchen is 303 K. Using 2159 J of work, how much heat can this refrigerator remove from its inside compartment

Answer 1

The heat is
`H = 23001.65 \ J`

Step-by-step explanation:

From the question we are told that

The inside temperature is
`T_i = 277 \ K`

Te kitchen temperature is
`T_k = 303 \ K`

The workdone is
`W = 2159 \ J`

The coefficient of performance of the refrigerator is mathematically represented as

`COP = (1)/((T_k)/(T_i) -1)`

substituting values

`COP = (1)/((303)/(277) -1)`

`COP = (277)/(26)`

`COP = 10.65`

Now this coefficient of performance of the refrigerator can also be represented mathematically as

`COP = (H)/(W)`

Where H is the heat which the refrigerator removes from the inside component

So

`H = COP * W`

substituting values

`H = 10.65 * 2159`

`H = 23001.65 \ J`