Heat Engine; Carnot Engine and refrigerator

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If minimum possible work is done by a refrigerator in converting 100 grams of water at $0^{0}$ C to ice, how much heat (in calories) is released to the surroundings at temperature $27^{0}$ C (Latent heat of ice = 80 cal/gram) to the nearest integer ?

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Solution

$\begin{array}{l} Q_{abs} = mL \\ Q_{abs} = 100 (80) cal \\ Q_{abs} = 8000 cal \\ N o w, CoP = \frac{Q_{abs}}{ω} = \frac{Q_{abs}}{Q_{rej} - Q_{abs}} = \frac{T_{s o u r c e}}{T_{\sin k} - T_{s o u r c e}} = \frac{273}{300 - 273} \\ \Rightarrow \frac{8000}{Q_{rej} - 8000} = 10.11 \\ \Rightarrow Q_{rej} = 8791.21 \\ \Rightarrow Q_{rej} \approx 8791 \end{array}$

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Heat Engine; Carnot Engine and refrigerator

Remember concepts with our Masterclasses.

If minimum possible work is done by a refrigerator in converting 100 grams of water at 00C to ice, how much heat (in calories) is released to the surroundings at temperature 270C (Latent heat of ice = 80 cal/gram) to the nearest integer ?

Qabs=mLQabs=100(80)calQabs=8000calNow, CoP=Qabsω=QabsQrej-Qabs=TsourceTsink-Tsource=273300-273⇒8000Qrej-8000=10.11⇒Qrej=8791.21⇒Qrej≈8791

Ready to Test Your Skills?

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If minimum possible work is done by a refrigerator in converting 100 grams of water at $0^{0}$ C to ice, how much heat (in calories) is released to the surroundings at temperature $27^{0}$ C (Latent heat of ice = 80 cal/gram) to the nearest integer ?