Question 7.2: Problem: The density of liquid water at atmospheric pressure...
Problem: The density of liquid water at atmospheric pressure and 0 °C is 1000 kg / m³ . When it freezes the density decreases by approximately 8%. How much would we have to increases the pressure to reduce the melting point of ice by 1 °C? The latent heat of fusion of ice is 334 kJ / kg.
Find: Increase in pressure ∆P to reduce melting point by 1 °C.
Known: Density of water at 0 °C ρ_f = 1000 kg / m³ , density of ice ρ_s is 8% less than ρ_f , latent heat of fusion of ice h_{sf} = 334 kJ / kg.
Assumptions: Densities ρ_s , ρ_f and latent heat h_{sf} are all constant.
Governing Equation:
Clapeyron equation \frac{dP}{dT} =\frac{h_{sg}}{T(v_g-v_s)}
Integrating the Clapeyron equation,
\int\limits_{P_1}^{P_2}{dP} =\int\limits_{T_1}^{T_2}{\frac{h_{sf}}{T(v_f-v_s)} } dT,
\Delta P =P_2-P_1 = \frac{h_{sf}}{(v_f-v_s)} \ln\frac{T_2}{T_1} .
Then knowing that
v_f = \frac{1}{\rho _f} =\frac{1}{1000 \ kg/m^3} =0.001 \ m^3/kg,
And the density of ice is 8% less than that of water:
\rho _s=(1-0.08) \times \rho _f=0.92 \times 1000 \ kg/m^3=920 \ kg/m^3,
v_s=\frac{1}{\rho _s} =\frac{1}{920 \ kg/m^3} =0.001 \ 087 \ 0 \ m^3/kg,
\Delta P = \frac{334 \ kJ/kg}{(0.001 \ m^3/kg – 0.001 \ 087 \ m^3/kg)} \ln\frac{272.15 \ K}{273.15 \ k} =14.081 \ MPa.
Answer: The pressure will have to be increased by 14.1 MPa to reduce the melting point by 1 °C.