# Adding heat to something without raising its temperature

I understand that you can add heat to something without raising its temperature during phase changes like from liquid to gas during boiling and solid to liquid during melting as the energy is used for the potential energy to separate the bonds between the molecules.

However, does it also apply during the phase change of liquid to solid during freezing where the energy is released instead of taken in? Will adding heat during the process of freezing also not raise its temperature?

Yes, what you say is true: if you add heat during the process of a liquid freezing into a solid, there will be no change in temperature. If two different phases are present in a system, then any addition or removal of heat will act to shift the phase balance one way or the other, rather than changing the temperature of the mixture (assuming the system reaches thermal equilibrium quickly).

So, in the example of water freezing, if you add some heat, some of the ice will melt back into water, but there will be no change in temperature of the system until it is either 100% water or 100% ice.

Heat added or extracted will not change the temperature if two phases are present and no work is being done. The heat transfer is given by

$$Q=mh$$

If we are dealing with a conversion between the solid and liquid phase (e.g.,ice and water at 0 C), $$h$$ is the latent heat of fusion, typically in units of $$\frac {kJ}{kg}$$. $$Q$$ is negative (heat out) when converting from liquid to solid and positive (heat in) when converting solid to liquid.

Once the phase conversion is completed continued addition or extraction of heat to or from the liquid or solid, respectively, will result in a temperature change of the liquid or solid according to

$$Q=mc\Delta T$$

Where $$c$$ is the applicable specific heat for the liquid or solid, typically in units of $$\frac {kJ}{kg^0C}$$.

$$\Delta T$$ is negative if heat is transferred out of the solid and positive if heat is transferred to the liquid.

Hope this helps.