To make ice, a freezer that is a reverse Carnot engine extracts 37 kJ as heat at -17°C during each cycle, with coefficient of performance 5.8. The room temperature is 26.9°C. How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?

Respuesta :

Answer:

a) 6.4 kJ

b) 43.4 kJ

Explanation:

a)

[tex]Q_{a}[/tex] = Heat absorbed = 37 kJ

[tex]β[/tex]  = Coefficient of performance = 5.8

[tex]W[/tex] = Work done

Heat absorbed is given as

[tex]Q_{a}[/tex] = [tex]β[/tex] [tex]W[/tex]

37 = (5.8) [tex]W[/tex]

[tex]W[/tex] = 6.4 kJ

b)

[tex]Q[/tex]  = work per cycle required

[tex]Q[/tex]  = [tex]Q_{a}[/tex] + [tex]W[/tex]

[tex]Q[/tex]  = 37 + 6.4

[tex]Q[/tex]  = 43.4 kJ

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