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

Answer:

The minimum workdone is    $W_d = 300 J$

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

                 $COP = \frac{T_2}{T_1 - T_2}$

    Substituting value

                 $COP = \frac{200}{400 - 200}$

                 $COP = 1$

This coefficient of performance can also be mathematically evaluated as

           $COP = \frac{E_b}{W_d}$

Where $W_d$ is the minimum workdone

    making $W_d$  the subject of the formula

             $W_d = COP * E_b$

So         $W_d =300 * 1$

            $W_d = 300 J$