A carnot refrigerator takes heat from water at 0 degree c and rejects heat to a room at 27 degree
c. if 100 kg of water at 0 degree c are to be changed to ice at 0 degree c, what is the required work in joules?
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The work required is 3313200 J.
To freeze the water at 0 degree we need to remove the Latent energy of fusion from the water, the value of latent heat is 335 kJ/kg
Therefore to freeze the water the amount of heat to be removed
Q=335*100=33500 kJ
Carnot COP of refrigerator is calculated as
COP=T_low/(T_high-T_low)=Heat removed/Work done
273/(300-273)=33500/W
W=3313.2 kJ=3313200 J
Therefore the work required in joules is 3313200 J.
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