Temperature is usually seen as a calibrated representation of heat but what about latent heat? Eg. Ice and water have different amounts of heat at 0 degree c.
Answer
Your statement:
Temperature is usually seen as a calibrated representation of heat
is at best only partially true. If the specific heat of your system is $C(T)$ then the heat put into your system in moving from temperature $T_1$ to $T_2$ is (assuming we can ignore work done on or by the system):
$$ U = \int_{T_1}^{T_2} \space C(T)dt $$
If the specific heat $C$ is independant of temperature then it's true that you get:
$$ U = C \space (T_2 - T_1) $$
and in this case you can regard the temperature as a "calibrated representation of heat". However this is a special case and in general $C(T)$ is a function of temperature.
As long as we stay away from a phase transition we expect $C(T)$ to be a smooth function of $T$ so there's no difficulty integrating to relate the heat change to the temperature change. However at a first order phase transition like melting or boiling the specific heat becomes singular and we can't just integrate through the phase transition. Instead if we have a phase transition between $T_1$ and $T_2$ we have to do something like:
$$ U = \int_{T_1}^{T_{phase}} C(T)dt + L + \int_{T_{phase}}^{T_2} C(T)dt$$
Where $L$ is the latent heat. You can think of this as $C(T)$ becoming a delta function at $T_{phase}$, and we can integrate delta functions to get a finite value, which in this case is the latent heat.
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