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

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-05-06T06:02:57+02:00

Answer:

The minimum workdone is [tex]W_d = 300 J[/tex]

Explanation:

From the question we are told that

The temperature range of the reservoir is [tex]T_1 \ to \ T_2 = 200K \ to \ 400K[/tex]

The energy to absorb is [tex]E_a = 300 J[/tex]

The coefficient of performance for the refrigerator is mathematically evaluated as

[tex]COP = \frac{T_2}{T_1 - T_2}[/tex]

Substituting value

[tex]COP = \frac{200}{400 - 200}[/tex]

[tex]COP = 1[/tex]

This coefficient of performance can also be mathematically evaluated as

[tex]COP = \frac{E_b}{W_d}[/tex]

Where [tex]W_d[/tex] is the minimum workdone

making [tex]W_d[/tex] the subject of the formula

[tex]W_d = COP * E_b[/tex]

So [tex]W_d =300 * 1[/tex]

[tex]W_d = 300 J[/tex]