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  • If minimum possible work is done by a refrigerator in converting grams of water at <img alt="0^{\circ}C" src="http://entrancecorner.codecogs.com/gif.latex?0%5E%7B

If minimum possible work is done by a refrigerator in converting 100 grams of water at 0^{\circ}C to ice, how much heat ( in calories) is released to the surroundings at temperature 27^{\circ}C (Latent heat of ice =80\; Cal/gram ) to the nearest integer ?
Option: 1 8791
Option: 2 309
Option: 3 4567
Option: 4 21

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\begin{array}{l} \mathrm{w}+\mathrm{Q}_{1}=\mathrm{Q}_{2} \\ \mathrm{w}=\mathrm{Q}_{2}-\mathrm{Q}_{1} \\ \\ \text { C.O.P. }=\frac{\mathrm{Q}_{1}}{\mathrm{w}}=\frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}-\mathrm{Q}_{1}}=\frac{273}{300-273}=\frac{\mathrm{Q}_{1}}{\mathrm{~W}} \\ \\ \mathrm{w}=\frac{27}{273} \times 80 \times 100 \times 4.2 \end{array}

\begin{array}{l} \mathrm{Q}_{2}=\mathrm{w}+Q_{1} \\ \\ \mathrm{Q}_{2}=(\frac{27}{273} \times 80 \times 100 \times 4.2)+(80 \times 100 \times 4.2) \\ \\ \mathrm{Q}_{2}=\frac{300}{273} \times 80 \times 100=8791.2 \mathrm{cal}\approx8791 \mathrm{cal} \end{array}

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