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A-refrigerator-use-1200joules-of-work-to-jump-300j-of-heat-from-a-cold-reservoir-of-275k-to-a-hot-reservoir-at-320k-1-What-is-the-coefficient-of-performance-2-the-maximum-efficiency-of-performan




Question Number 179289 by Tawa11 last updated on 27/Oct/22
A refrigerator use 1200joules  of work to jump 300j of heat from a cold reservoir of 275k to a hot reservoir at 320k.    1. What is the coefficient of performance.   2 the maximum efficiency of performance.
A refrigerator use 1200joules of work to jump 300j of heat from a cold reservoir of 275k to a hot reservoir at 320k.

1. What is the coefficient of performance.
2 the maximum efficiency of performance.

Answered by Acem last updated on 27/Oct/22
1• K= ((∣Q_c ∣)/(∣W∣))= ((300)/(1200))= (1/4)  2•  K_(carnot) = (T_c /(T_H −T_c )) = 6.1
$$\mathrm{1}\bullet\:{K}=\:\frac{\mid{Q}_{{c}} \mid}{\mid{W}\mid}=\:\frac{\mathrm{300}}{\mathrm{1200}}=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{2}\bullet\:\:{K}_{{carnot}} =\:\frac{{T}_{{c}} }{{T}_{{H}} −{T}_{{c}} }\:=\:\mathrm{6}.\mathrm{1} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 28/Oct/22
I appreciate sir.
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Commented by Acem last updated on 28/Oct/22
You′re welcome
$${You}'{re}\:{welcome} \\ $$

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