What volume of water at 0 ∘C∘C can a freezer make into ice cubes in 2.0 hh, if the coefficient of performance of the cooling unit is 9.0 and the power input is 1.2 kilowatt?

19170 cm³

Explanation:

The formula of coefficient of performance is given as,

β = T₂ / (T₁-T₂) ... Equation 1

Where T₂ = Final Temperature, T₁ = initial Temperature, β = coefficient of performance.

Given: β = 9.0, T₁ = 0 °C = 273 K.

Substitute into equation 1

9 = T₂ / (273-T₂)

T₂ = 2457-9T₂

10T₂ = 2457

T₂ = 2457/10

T₂ = 245.7 K

T₂ = - 27.3 °C.

From specific heat capacity,

Q = lm + cm (T₁-T₂) ... Equation 2

Where Q = Amount of heat, l = specific latent heat of water, m = mass of water, c = specific heat capacity of water.

But,

Q = Pt

where P = power input, t = time.

P = 1.2 kW = 1200 W, t = 2 h = 7200 s

Q = 1200*7200

Q = 8640000 J.

Given: T₁ = 0 °C, T₂ = - 27.3 °C

Constant: c = 4200 J/kg. K, l = 336000 J/kg

Substitute into equation 2

8640000 = 336000m + 4200 (m) [0 - (-27.3) ]

8640000 = 336000m + 114660m

8640000 = 450660m

m = 8640000/450660

m = 19.17 kg.

From density,

Volume = Mass/Density.

V = M/D

Where M = mass of water, D = Density of water

Given: M = 19.17 kg, D = 1000 kg/m³

V = 19.17/1000

V = 0.01917 m³

V = 19170 cm³

Hence the volume of water = 19170 cm³