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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.

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Related: physics.stackexchange.com/questions/41610/… –  John Rennie Feb 12 '13 at 16:04
    
For one thing, heat and temperature have distinct units of measurement, the former being expressed in SI units of Joules, resp. Kelvin. Think of heat as an extensive quantity and temperature as intensive quantity. –  alancalvitti Feb 18 '13 at 6:09

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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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There is no thermodynamic state variable called "heat". The only way in which it makes sense to talk about "heat" is in the differential way, $\delta Q$, the infinitesimal heat added or taken out of a system during some process. This, it turns out, is not an exact differential, and this means that the integral over $\delta Q$ from some initial to some final state point depends on the path taken, and not just on the end points (which is exactly why no state function "heat" exists). However, inexact differentials can be turned into exact differentials by integrating factors, and the integrating factor which makes the differential of heat exact is the inverse temperature: $\frac{1}{T}\delta Q$ is exact. In fact, it is the differential of the entropy, $dS$.

I doubt that this is what you wanted to hear. But then, the concept of heat is often so much confused, that I felt it necessary to spell out some pedantic facts.

If I slightly reword your question and interpret it as "what is the difference between energy and temperature", then I can say a bit more. If you have two systems which are brought in contact in such a way that they can exchange energy, then ultimately they will reach the same temperature. This does not mean that they will have the same energy, though, because one of the two systems could be much bigger than the other.

In a famous analogy that goes back to Feynman, think of the process of drying yourself with a towel when you're wet. The towel takes up water from you and gets wetter, while you get dryer. But if the towel is already pretty wet, you will not be able to completely dry yourself with it, since water will get from you onto the towel as much as water will get from the towel onto you. In this "equilibrium" situation the "wetness" of you and of the towel are in equilibrium, but that does not mean that the amount of water on the towel is the same as the amount of water on you. Now think "water"="energy" and "wetness"="temperature".

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Temperature is a physical quantity that measures the hotness or coldness of a body but heat is a form of energy which causes the sensation of warm.

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Then you may want to define hotness and coldness more accurate. –  Ali Jul 24 '13 at 13:54

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