Assuming a domestic refrigerator as a reversible heat engine working between melting point
Assuming a domestic refrigerator as a reversible heat engine working between melting point Of ice and the room temperature at 27°C, calculate the energy in joule that must be supplied to freeze 1Kg of water at 0°C.
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Here T1 = 27+273 =300K , T2 =0 +273 = 273
Mass of water to be freezed , M = 1 Kg = 1000g
Amount of heat that should be removed to freeze the water
Q2 = ML =1000X 80 cal
= 1000X80 X4.2 =3.36 x 105 J
Now Q1 = (T1/T2 )X Q2 = (300/273)X3.36x105 = 3.692 x105 J
Therefore energy supplied to freeze the water
W =Q1 – Q2 = 3.693x105 - 3.36 x105
= 3.32 x105 J
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