Assuming a domestic refrigerator as a reversible heat engine working b

1.	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.
Answer» 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

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.

Answer»

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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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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