A plant produces ice cubes. 0.45 kg/min of liquid water at 0 o C enters a coolin

Question

A plant produces ice cubes. 0.45 kg/min of liquid water at 0 o C enters a cooling chamber where it is transformed into ice at 0 o C. The cooling chamber is directly connected to a Carnot heat pump (powered by a 186 W electrical mo tor) that transf er to the surroundings th e heat given off during the freezing process. What is th e temperature of the surroundings? Assumptions: Ice naturally flows out of the cooling chamber. The cooling chamber is perfectly insulate d except its connection to the heat pump.

Explanation / Answer

Enthalpy of water at 0 o C = 2500.9 Kj /kg

water is produced at rate of 0.45 kg /min = 0.45 /60 kg /sec

so power transferred = 0.45/60 * 2500.9 Kj /sec = 18.756 Kw .

cop (coefficient of perormance ) = Qh/W = Th / ( Th - Tl ) ...... (1)  where Tl ,Th are lower and higher temperature.

here W = 186 W , Qc = 18.756Kw Qh = Qc + W ,

Tl = 273 K Th = 275.7 K by using above equation (1) .

Hence the surrounding temperature is 275.7 K.

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