A Carnot's engine works as a refrigerator bet | vedclass

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Question

(A) Given: Temperature of the cold reservoir,$T_2 = 250\, K$. Temperature of the hot reservoir,$T_1 = 300\, K$. Heat extracted from the cold reservoir

Vedclass · Accepted Answer

$420$

Vedclass · Answer

(A) Given: Temperature of the cold reservoir,$T_2 = 250\, K$. Temperature of the hot reservoir,$T_1 = 300\, K$. Heat extracted from the cold reservoir,$Q_2 = 500\, cal$. For a Carnot refrigerator,the coefficient of performance $\beta$ is given by $\beta = \frac{T_2}{T_1 - T_2} = \frac{Q_2}{W}$. Substituting the values: $\beta = \frac{250}{300 - 250} = \frac{250}{50} = 5$. Now,using $\beta = \frac{Q_2}{W}$,we get $W = \frac{Q_2}{\beta} = \frac{500\, cal}{5} = 100\, cal$. Since $1\, cal = 4.2\, J$,the work done in Joules is $W = 100 	imes 4.2\, J = 420\, J$.

$A$ Carnot's engine works as a refrigerator between $250\, K$ and $300\, K$. It receives $500\, cal$ heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is ..... $J$

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