Why is the latent heat of fusion of helium-3 a negative quantity? I am new to this community and I have little to no background in physics, so I apologise should I make a glaring factual mistake in my question. 
I briefly studied the phase transitions of matter and, quite by accident, came across the "latent heat of fusion", defined as the energy needed to be supplied to the system to change it from solid to liquid. It reads on Wikipedia that this quantity is, naturally, generally positive as an increase in internal energy is needed for the transition. However, there is apparently one notable exception as is the case with helium. 

Quote: "Helium-3 has a negative enthalpy of fusion at temperatures below 0.3 K. [...] This means that, at appropriate constant pressures, these substances freeze with the addition of heat."

Could anyone kindly explain to me, in laymen terms please, why is it the case that when heat is added to this substance, it FREEZES at constant pressure? It honestly fascinates me and I can't wrap my head around it. 
 A: I have attempted to answer this interesting question noting the following:

I have little to no background in physics
Could anyone kindly explain to me, in laymen terms please

The behaviour of $^3 \rm He$ is anomalous if not unique because of a number of reasons.
The first reason is that the interaction between $^3 \rm He$ atoms is very weak so much so that it is only at temperature below $3.19\,\rm K \approx -270\, ^\circ \rm C$ that you can liquify $^3 \rm He $.
The second reason is that in the nucleus of $^3 \rm He $ there are two protons and one neutron each of which have a property called spin with a value of $\frac 12$ and overall the $^3 \rm He $ nucleus has a total spin of $\frac 12$.
An important physical quantity which I will use is entropy which you can think of as a measure of disorder and if you add heat to a system you will increase the disorder and hence the entropy.
So you can melt solid water crystals (ice) by adding heat to it to form liquid water in which the molecules are less ordered thus increasing the entropy of the water.
A useful visual aid which is used when discussing the phases of a substance is called a phase diagram which is a graph of pressure against volume.
A normally behaved substance has a phase diagram like the one below which is for carbon dioxide.

Suppose you have some carbon dioxide which is at a temperature of $-78.5\, ^\circ \rm C$ and a pressure of $30$ atmospheres (C) you will see from the graph that the carbon dioxide is a solid.
Now  keeping the pressure constant at 30 atmospheres consider what would happen of the carbon dioxide was heated from a temperature of $-78.5\, ^\circ \rm C$ (C) to a temperature of $20 \, ^\circ \rm C$ (B)?
At first the temperature would rise and the carbon dioxide would be in the solid state until the temperature reached approximately $-55\, ^\circ \rm C$.
The temperature would stay constant and the carbon dioxide would start to melt transitioning from the solid state to the liquid state.
The heat which must be added for this transition to occur is called the enthalpy of fusion and it is a positive quantity.
Once all the solid carbon dioxide becomes liquid its temperature will start to increase and at a temperature of $20\, ^\circ \rm C$ (B) the carbon dioxide is a liquid.
All the time the temperature of the carbon dioxide has increased the carbon dioxide has become more disordered and its entropy has increased.
If the pressure is now reduced to $1$ atmosphere and the temperature is kept constant at $20\, ^\circ \rm C$ the carbon dioxide will undergo another phase transition from liquid to gas and at $1$ atmosphere and $20\, ^\circ \rm C$ carbon dioxide is a gas.
The phase diagram for $^3 \rm He $ is very different from that of carbon dioxide.

You will note that unless the pressure is greater than $29$ atmosphere  $^3 \rm He $ cannot be in the solid state.
This is because to the very weak interaction between the atoms of $^3 \rm He $ which have to be squeezed closer together to form the solid state.
With $^3 \rm He $ at $30$ atmospheres and at about $0.1 \rm K$ it is a liquid (D).
If the temperature of the liquid $^3 \rm He $ is allowed to rise there will come a temperature when it starts to become a solid and given enough time it will be all solid (E).
Allowing the temperature to rise further the $^3 \rm He $ will undergo another phase transition from solid to liquid eventually the $^3 \rm He $ will all be in the liquid phase.
The transition from liquid to solid when the temperature is rising is equivalent to the enthalpy of fusion being negative.
To explain this behaviour  one must turn to quantum mechanics.
It would seem that during that transition one goes from a less disordered liquid to a more disordered solid and that is in fact what is actually happening.
This is where the spin $\frac 12$ of the $^3 \rm He $ nucleus becomes important.
A visual aid for spin is to imagine that each nucleus is an arrow and when there is disorder a collection of nuclei have all the arrows pointing in random directions.
The most ordered state is when the arrows all point in the same direction.
The disordered state (random direction of spins) has a larger entropy than the ordered state (spins all pointing in the same direction).
As the temperature of liquid $^3 \rm He $ decreases so does the spins become more ordered (arrows pointing in the same direction) and so the entropy of the liquid state decreases with temperature.

However in the solid state the nuclei stay disordered and the entropy does not changed with temperature.
You will note from the graph of entropy against temperature there is a cross over point at a temperature of $0.3\,\rm K$ below which the entropy and hence disorder of liquid $^3 \rm He $ is less than that of solid $^3 \rm He $.
So there is this very strange condition for $^3 \rm He $ where the solid state is less ordered than the liquid state which is exactly the opposite for carbon dioxide and all? other substances except $^4 \rm He $.
This anomalous behaviour of $^3 \rm He $ can be used as a basis for cooling below $0.3 \rm K$ down to about $1 \,\rm mK$ because if one increases the pressure on a sample of liquid $^3 \rm He $ such that it turns liquid to  solid, this requires an input of energy which can come from the thermal energy of the sample itself - the temperature of the sample decreases.
This is called Pomeranchuk cooling and was used by Lee, Osheroff and Richardson in their Nobel prize (1996) winning discovery of the superfluidity of liquid $^3 \rm He $.
