# Changing the melting point of ice

Myself and a friend have been stuck on this problem for a while now. We know that we should end up with an answer of $-0.007K$ however, cannot get to this solution. Help with the method we should use would be great. Thanks.

Estimate the change in the melting temperature of ice, initially at standard temperature and pressure (s.t.p.), when the pressure is increased by one atmosphere ($1.01 × 10^5 Pa$). The density of water at s.t.p. is $1\times10^3\ kg\ m^{-3}$, the density of ice at s.t.p. is $917\ kg\ m^{−3}$ and the latent heat of fusion of ice per unit of mass at s.t.p is $3.35\times 10^5\ J\ kg^{−1}$.

Currently we have tried two possible ways, but none have worked.

That is:

Solving the differential equation: $\frac{dP}{dT}=\frac{L}{T(V_2-V_1)}$

But this did not seem to work, so we the tried to use: $ln(\frac{P_2}{P_1})=-\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

I managed to figure this out!

It was done using the Clausius Clapeyron equation.

$$\frac{dP}{dT}=\frac{L}{T(V_2-V_1)}$$

By letting $L=\frac{h_f}{m}$ and $V_1=\frac{1}{\rho_1}$,$V_2=\frac{1}{\rho_2}$ gives:

$$\frac{dP}{dT}=\frac{\frac{h_f}{m}}{T(\frac{1}{\rho_2}-\frac{1}{\rho_1})}$$

By solving this differential equation it is seen that:

$$\int_{P_1}^{P_2}dP=\frac{\frac{h_f}{m}}{\frac{1}{\rho_2}-\frac{1}{\rho_1}}\int_{T_1}^{T_2}\frac{1}{T}dT$$

Therefore:

$$P_2-P_1=\frac{\frac{h_f}{m}}{\frac{1}{\rho_2}-\frac{1}{\rho_1}}(ln\left(\frac{T_2}{T_1}\right))$$

Now, substituting in the values gives $T_2=293.007K$

So the change in melting point would be $$T_2-T_1=0.007K$$ as was required. $\square$