Question

a Carnot engine extracts heat from water at 0 degree centigrade and rejects it into room at 24.4degree centigrade. the work required by the refrigerator for every 1kg of water converted into ice (latent heat of ice=336kj/kg)is

Answer 1

Answer:

Work required to be done by the refrigerator is 27.55 kJ.

Explanation:

Given data:

The absolute temperature of the cold refrigerator, Tc = 0°∁ = 273 K

The absolute temperature of the hot refrigerator, Th = 24.4°∁ = 24.4 + 273 K = 297.4 K

1 kg of water is converted into ice

Latent heat of ice = 336 kJ/kg

To find: work required a refrigerator to convert 1 kg of water into ice

The efficiency of the Carnot refrigerator is given by,

ɳ = (Th – Tc) / Th  

ɳ = (297.4–273) / 297.4 = 24.4 / 297.4 = 0.082  

Now,  

the heat required to convert the 1 kg of water into ice, Q  

= 1 * 336 kJ/kg

= 336kJ/kg

We know, in Carnot Cycle the work  done by the refrigerator is given by

W = ɳ * Q

∴ W = 0.082 * 336 kJ/kg = 27.55 kJ

Answer 2

Explanation:

answer is perfectly alright