Question

A freezer has a coefficient of performance of 2.60. The freezer is to convert 1.86 kg of water at 26.0 ∘C to 1.86 kg of ice at -5.00 ∘C in 1 hour.

a) What amount of heat must be removed from the water at 26.0 ∘C to convert it to ice at -5.00 ∘C ?

b) How much electrical energy is consumed by the freezer during this hour?

c) How much wasted heat is rejected to the room in which the freezer sits?

Answer #1

give that

mass of water m= 1.86 kg

cofficient of performance k = 2.60

initial temperature of water T0 = 26 degree C = 299 K

final temperature of ice is T1 = - 5 degree C = 268 K

(a)

total amount of heat must be removed is when

water at 26 degree C converted to water at 0 degree C

Q1 = m*c_{V}*T

Q1 = 1.86*4190*(299-273)

Q1 = 2.02*10^5 J

amout of heat when 0 degreeC water converted to ice 0 degree C

Q2 = m*Lf = 1.86*334*10^3 = 6.21*10^5 J

amount of heat when ice at 0 degree C converted to ice at -5 degree C

Q3 = m*c_{ice}*T

Q3 = 1.86*2100(273-268) = 1.95*10^4 J

total amount of heat removed is

Q = Q1 + Q2 + Q3 = 2.02*10^5 + 6.21*10^5 + 1.95*10^4

Q = 8.42*10^5 J

(b)

the electric energy consumed is

W = Qc /K

W = 8.42*10^5 / 2.60 = 3.24*10^5 J

(c)

the wasted heat delivered is

Q_{H} = Qc + H

Q_{H} = 8.42*10^5 + 3.24*10^5

Q_{H} = 11.66 *10^5 J

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