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?
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*cV*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*cice*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
QH = Qc + H
QH = 8.42*10^5 + 3.24*10^5
QH = 11.66 *10^5 J
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