Melting of substance with -ve slope for fusion curve

[Parzeevahl]

Homework Statement

Show that a substance with a -ve slope for its fusion curve, such as ordinary ice or bismuth, contracts upon melting.

Relevant Equations

Clausius-Clapeyron equation:
Delta T (change in temp.) = (T * (v(f) - v(i)) * Delta P) / (h(f) - h(i))
where 'Delta P' is the change in pressure.

The question says that the process is melting, so temperature must increase.
Hence, Delta T > 0.
Also, it is given that the slope for its fusion curve is -ve, which means that as we increase temperature, the pressure will decrease.
So, Delta P < 0.
The question asks to prove that the substance contracts upon melting, i.e., Delta v (change in vol.) < 0.
So, in the right-hand side of the Clausius-Clapeyron equation, for both Delta P and Delta v = (v(f) - v(i)) to be -ve, (T / (h(f) - h(i))) must be +ve, as in the left-hand side, Delta T > 0. So essentially the problem comes down to proving that (T / (h(f) - h(i))) > 0.
Now, we are using absolute temperature here, T is always greater than or equal to zero.
So, my question is, how do I prove that h(f) - h(i), i.e., the change in enthalpy, is +ve in this melting process?

 

[Chestermiller]

Problem Statement: Show that a substance with a -ve slope for its fusion curve, such as ordinary ice or bismuth, contracts upon melting.
Relevant Equations: Clausius-Clapeyron equation:
Delta T (change in temp.) = (T * (v(f) - v(i)) * Delta P) / (h(f) - h(i))
where 'Delta P' is the change in pressure.

The question says that the process is melting, so temperature must increase.
Hence, Delta T > 0.
Also, it is given that the slope for its fusion curve is -ve, which means that as we increase temperature, the pressure will decrease.
So, Delta P < 0.
The question asks to prove that the substance contracts upon melting, i.e., Delta v (change in vol.) < 0.
So, in the right-hand side of the Clausius-Clapeyron equation, for both Delta P and Delta v = (v(f) - v(i)) to be -ve, (T / (h(f) - h(i))) must be +ve, as in the left-hand side, Delta T > 0. So essentially the problem comes down to proving that (T / (h(f) - h(i))) > 0.
Now, we are using absolute temperature here, T is always greater than or equal to zero.
So, my question is, how do I prove that h(f) - h(i), i.e., the change in enthalpy, is +ve in this melting process?

You have to add heat to get something to melt, so the change in enthalpy in going from solid to liquid is positive.