Question

A carnot freezer takes heat from water at 0 c inside it and rejects it to the room at a temperature of 27 c.The latent heat of ice is 336 x 10^3.If 5 kg of water at 0 c is converted into ice at 0 c by the freezer consummed by the freezer is close to

Answer 1

Hey dear,

◆ Answer-
E = 166.4 kJ

◆ Explaination-
# Given-
m = 5 kg
lh = 36×10^3 J/kg
T1 = 0 °C = 273 K
T2 = 27 °C = 300 K

# Solution-
Heat required to freeze given ice -
Heat required = mass × latent heat
Q1 = m × lh
Q1 = 5 × 336×10^3
Q1 = 1.68×10^6 J

In Carnot cycle -
Q ∝ T
Therefore,
Q1 / Q2 = T1 / T2
Q2 = Q1 × T2 / T1
Q2 = 1.68×10^6 × 300 / 273
Q2 = 1.846×10^6 J

Energy consumed by the freezer is calculated by -
E = Q2 - Q1
E = 1.846×10^6 - 1.680×10^6
E = 0.1664×10^6 J
E = 166.4 kJ

Energy consumed by the freezer is 166.4 kJ.

Hope this helps you...