Physics, asked by newtismypotato9430, 1 year ago

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?

Answers

Answered by Ursus
7

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.

Similar questions