Answer:
19170 cm³
Step-by-step 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