The temperature inside and outside of a refrigerator are $275$ K and $300$ K, respectively. Assumng that the refrigerator cycle is ideal reversible, for every joule of work done the heat delivered to the surroundings is nearly.

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Solution

The correct option is B $11$ J

$B = \frac{Q_{2}}{W}$

$B = \frac{1}{n} - 1$
= $\frac{T_{2}}{T_{1} - T_{2}}$

= $\frac{273}{27}$ = 10.1

$10.1 = \frac{Q_{2}}{W}$

$Q_{2} = 10.1$

$Q_{1} = Q_{2} + W$

= 10.1 + 1 = 11.1

$Q_{1} \approx 11 J$

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The temperature inside and outside of a refrigerator are 275K and 300K, respectively. Assumng that the refrigerator cycle is ideal reversible, for every joule of work done the heat delivered to the surroundings is nearly.

The correct option is B 11JB=Q2WB=1n−1= T2T1−T2 = 27327 = 10.110.1=Q2WQ2=10.1Q1=Q2+W = 10.1 + 1 = 11.1Q1≈11J

The temperature inside and outside of a refrigerator are $275$ K and $300$ K, respectively. Assumng that the refrigerator cycle is ideal reversible, for every joule of work done the heat delivered to the surroundings is nearly.

The correct option is B $11$ J

$B = \frac{Q_{2}}{W}$

$B = \frac{1}{n} - 1$
= $\frac{T_{2}}{T_{1} - T_{2}}$

= $\frac{273}{27}$ = 10.1

$10.1 = \frac{Q_{2}}{W}$

$Q_{2} = 10.1$

$Q_{1} = Q_{2} + W$

= 10.1 + 1 = 11.1

$Q_{1} \approx 11 J$