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 surrounding at temperature \(27^{\circ} C\) (Latent heat of ice \(=80 Cal / gram\) ) to the nearest integer?

Pick an option to check your answer.

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Answer & Solution

Correct Answer

(C) \(8791\)

Step-by-step Solution

Detailed explanation

\(W + Q _{1}= Q _{2}\)

\(W = Q _{2}- Q _{1}\)

\(C.O.P. =\frac{ Q _{1}}{ W }=\frac{ Q _{1}}{ Q _{2}- Q _{1}}=\frac{273}{300-273}=\frac{ Q _{1}}{ W }\)

\(W =\frac{27}{273} \times 80 \times 100 \times 4.2\)

\(Q _{2}= W+Q_{1}\)

\(Q _{2}=\frac{27}{273} \times 80 \times 100 \times 4.2+80 \times 100 \times 4.2\)

\(Q _{2}=\frac{300}{273} \times 80 \times 100=8791.2 cal\)

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