Confusion regarding latent heat of fusion

Question

During vaporizing there is higher increase in internal energy (higher positive $\Delta U$) and more work is done by the liquid (higher $W$) as molecules become widely separated.

During melting, there is small increase in internal energy (smaller positive $\Delta U$) and less work is done by the solid (smaller $W$) as there is less difference in the molecule separation relatively.

According to $Q=\Delta U - W$, why is the specific latent heat of vaporization greater than that of fusion? In both cases $Q$ works out to be same according to above statements?

Answer 1

Let me clear a few things up first; the latent heat of fusion is the energy required to convert a substance from solid form to a liquid form. Since water is liquid at room temperature, the latent heat of fusion is positive as energy is absorbed to convert ice to water, just as energy is released when water is converted to ice. It sounds counter-intuitive, I know, but release could also mean 'sucked away', which is exactly what is done in a freezer.

Now, molecules close together in solids, a little less close together in liquids, and extremely far apart in gases. There is a lot of cohesion in both ice and liquid water, however very little in water vapor. The great difference in energy required to melt ice and to form water vapor is because a lot of energy is required to overcome cohesion.

This cohesion occurs because of the polarity of water molecules. This video should help you out!

Answer 2

I was simply confused by the sign convention in the first law of thermodynamics.

The most convenient way to write it is $Q=\Delta U + W$ where $W$ is the work done by the system.

Hence the answer to my question becomes clear after this.

Answer 3

It should be (delta)U=q+-w, where +- means plus or minus, like the sign used in the quadratic formula. The sign for w should be determine by the type of work done, ie. compression (+w) or expansion work (-w).