# Why the units of heat of fusion and heat of evaporation don't include the factor of temperature?

I know that the units of specific heat are:

$$\frac{J}{kg\cdot °C}$$

On the other hand, the units of heat of fusion $L_{f}$ and heat of evaporation $L_{v}$ are

$$\frac{J}{kg}$$

So, why don't they include the factor $°C$?

• Because when there is a change of state the temperature does not change. – Farcher Jan 29 '17 at 17:13
• @Farcher : Brief as it is, this is an answer not a comment . As far as I am aware there are no rules prohibiting short answers. – sammy gerbil Jan 29 '17 at 17:50
• @sammygerbil Thanks for the advice. One day I will get the hang of what one can do and what one cannot. :-) – Farcher Jan 29 '17 at 19:43

Because they are energies, and energy is not necessarily tied to temperature.

Latent heats $L_f$ and $L_v$ are the amounts of energy you need to melt/solidify or vaporize/condense something - and this could happen at any temperature depending on other parameters of the materials.

Don't be confused about the word "heat". It has not got more to do with temperature than e.g. work.

Example

Consider an ice cube at $0\;^\circ \mathrm C$. To make it melt, you try to heat it up. But it doesn't heat up before all ice is melted. The amount of heat added is only used for the melting, not for raising the temperature

If I add just a tiny amount of heat, far from enough to melt all the ice, then the temperature can't possibly rise. If it rose, then it would be higher than $0\;^\circ \mathrm C$, which is impossible for ice. So therefore, adding this tiny amount of heat doesn't raise the temperature. So what does it do? It melts some of the ice instead (the energy has to be used for something).

If $L_f=400\;\mathrm{kJ/kg}$ and I try to melt a $2\;\mathrm{kg}$ ice block then we need $800\;\mathrm{kJ}$ before all is melted. During melting, no temperature change - because such temperature change would spend some of this added energy, and then there wouldn't be enough left to melt the ice - and then all the ice wouldn't be melted, and then a temperature change to more than $0\;^\circ \mathrm C$ is impossible. But if I add $850\;\mathrm{kJ}$, then the ice will melt and then the temperature will rise because of the excess $50\;\mathrm{kJ}$.

It seems to me that you are seeing temperature as something that must be influenced by heat. But this is not the case. There are many other ways energy can influence an object - in the case of melting, it changes it's phase (from solid to liquid). Phase changes and temperature changes are side by side - one is not more superior than the other. Both need energy, and if one takes all, the other won't change.

• Could illustrate what you say in the second paragraph, for example with a ice cube? Thanks. – InfZero Jan 29 '17 at 17:16
• @InfZero I have added a description of an ice cube. Does this clear out some confusion? – Steeven Jan 29 '17 at 17:34

When heat flows into ice whose temperature is lower than the freeing point, the consequence is that the particles vibrate faster. That is, there is an increase in kinetic energy of the particles. Temperature is basically defined as the measure of the average KE of the particles in a substance. The particles of the warmer ice would impart energy to the liquid in a traditional thermometer causing the fluid to expand and rise in the thermometer. The same is true of heating liquid water.

But when heat flow into ice at 0 centigrade, the energy now goes into pulling the particles from their previous positions. Let me switch to boiling for a moment, it's easier to visualize. The molecules in liquid water are kept in contact by electrical attractions, hydrogen bonds here, between the molecules. The energy in the additional input of heat while boiling goes not into making the molecules move faster but into pulling them apart from one another. This is a form of "potential energy" as opposed to kinetic energy. This is comparable to lifting a mass higher above the Earth. Since there is no increase in KE during boiling, there is no increase in temperature. The same is true during melting, the energy in added heat ends up as potential energy versus kinetic energy.

If a glass of ice water is well mixed, it will be at 0 Celcius regardless of the proportion of liquid and solid.