JEE Mains · Physics · STD 11 - 11. thermodynamics
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?
- A \(8000\)
- B \(8502\)
- C \(8791\)
- D \(8561\)
Answer & Solution
Correct Answer
(C) \(8791\)
Step-by-step Solution
Detailed explanation
\(W + Q _{1}= Q _{2}\) \(W = Q _{2}- Q _{1}\) \(C.O.P. =\frac{ Q _{1}}{ W }=\frac{ Q _{1}}{ Q _{2}- Q _{1}}=\frac{273}{300-273}=\frac{ Q _{1}}{ W }\) \(W =\frac{27}{273} \times 80 \times 100 \times 4.2\) \(Q _{2}= W+Q_{1}\)…
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