Question

You wish to buy a new freezer for your basement. The primary use of the freezer will be to prepare ice for family parties. You have a very large family, with grandparents, parents, sisters, brothers, uncles, aunts, cousins, nephews, nieces, and grandchildren. As a result, family birthday parties are held every couple of weeks, and you need lots of ice for drinks. You want the freezer to convert 10.1 kg of water at 23.0°C to 10.1 kg of ice at

−9.4°C

in 2.00 h. But, in order to keep your electric bill down, you want the power rating of the freezer to stay below 100 W. From these requirements, you determine the minimum COP of the freezer that will satisfy your needs. (Consider that the specific heat of water is 4,186 J/(kg · °C), the specific heat of ice is 2,090 J/(kg · °C), the latent heat of fusion of water is 3.33 ✕ 10⁵ J/kg, and the latent heat of vaporization of water is 2.26 ✕ 10⁶ J/kg.)

Answer 1

Specific heat of water = C₁ = 4186 J/(kg.^oC)

Specific heat of ice = C₂ = 2090 J/(kg.^oC)

Latent heat of fusion of water = L₁ = 3.33 x 10⁵ J/kg

Latent heat of vaporization of water = L₂ = 2.26 x 10⁶ J/kg

Mass of water to be converted into ice in the freezer = m = 10.1 kg

Initial temperature of water = T₁ = 23 ^oC

Freezing point of water = T₂ = 0 ^oC

Final temperature of ice = T₃ = -9.4 ^oC

Heat energy to be removed from the water = H

H = mC₁(T₁ - T₂) + mL₁ + mC₂(T₂ - T₃)

H = (10.1)(4186)(23 - 0) + (10.1)(3.33x10⁵) + (10.1)(2090)(0 - (-9.4))

H = 4.534 x 10⁶ J

Time period to turn the water into ice at -9.4 ^oC = t = 2 hours = 2 x (3600) sec = 7200 sec

Rate at which heat is removed from the water = Q

H = Qt

4.534x10⁶ = Q(7200)

Q = 630 W

Power rating of the freezer = W = 100 W

Coefficient of performance of the freezer = COP

COP = 6.3

Minimum COP of the freezer that will satisfy your needs = 6.3