A carnot refrigeration cycle absorbs heat at $270 K$ and rejects heat at $300 K$ . If the cycle is absorbing $1260 K J / m i n$ at $270 K$ , then work required per second is :

2.33 K J / s e c .

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4.66 K J / s e c .

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1 K J / s e c .

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4 K J / s e c .

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Solution

The correct option is A $2.33 K J / s e c .$
Given that,

Temperature $T_{1} = 270 K$

Temperature $T_{2} = 300 K$

Absorbed heat $Q = 1260 K J / m i n = 21 K J / s$

We know that,

Coefficient of performance is

C.O.P = $\frac{T_{2}}{T_{2} - T_{1}}$

C.O.P= $\frac{270}{300 - 270}$

C.O.P=9

Using formula of work done

C.O.P= $\frac{Q}{W}$

$W = \frac{21}{9}$

$W = 2.33 K J / s$
The work required is $2.33 K J / s$ .

Hence, This is the required solution.

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