Question

A refrigerator is a heat engine running in reverse. Work that is done on the refrigerator enables it to absorb heat from a low temperature reservoir and release heat to a high temperature reservoir. Consider an ideal refrigerator connected to a 200 K and 400 K reservoir. What is the minimum amount of work (in J) required for this refrigerator to absorb 300 J of energy from the low temperature reservoir

Answer 1

The minimum workdone is
`W_d = 300 J`

Step-by-step explanation:

From the question we are told that

The temperature range of the reservoir is
`T_1 \ to \ T_2 = 200K \ to \ 400K`

The energy to absorb is
`E_a = 300 J`

The coefficient of performance for the refrigerator is mathematically evaluated as

`COP = (T_2)/(T_1 - T_2)`

Substituting value

`COP = (200)/(400 - 200)`

`COP = 1`

This coefficient of performance can also be mathematically evaluated as

`COP = (E_b)/(W_d)`

Where
`W_d` is the minimum workdone

making
`W_d` the subject of the formula

`W_d = COP * E_b`

So
`W_d =300 * 1`

`W_d = 300 J`