Question

the temperature inside and outside a refrigerator are 273k and 300k respectively.assuming that the refrigerator cycle is reversible,for every joule of work done,the heat delivered to the surrounding will be nearly what?

Answer 1

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Answer 2

Answer: 11 J

Explanation:

The heat delivered to the surroundings = Q₁

For cyclic refrigerator:

$\frac{T_2}{T_1-T_2} = \frac{Q_2}{W}$

$W =Q_1-Q_2\\T_2 = 273 K\\T_1 = 300 K$

W = 1 J

$\frac{273}{300-273} = \frac{Q_2}{1 J}\Rightarrow Q_2=10.11 J$

$Q_1 = Q_2+W = 10.11 +1 =11.11 J = 11 J$