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Is the rate of change in temperature for an object constant? For example, from 0ºC to 25ºC, or from 25ºC to -10ºC? Does it take the same amount of time to increase temperature from 1º to 2º as 24º to 25º, or 0 to -1 as -9 to -10.

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In general, no. But maybe constant rate of temperature change is a reasonable assumption for the situation you have in mind. You might want to add more information about what type of object and circumstances you are thinking about. Right now the question is very vague. – jkej Feb 4 '13 at 12:44
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I think you may be interested in Newton's Law of Cooling. This law states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings. For example, if the atmosphere is of temperature $T_{atm}$ and our object is of a different temperature $T$, we can represent Newton's law of cooling as:

$\frac{dT}{dt} = k(T_{atm} - T)$

where $t$ represents time, and $k$ is a constant.

Re-arranging this equation and integrating both sides gives:

$\int\frac{dT}{(T_{atm} - T)} = \int kdt$

which evaluates to:

$-\ln{(T_{atm} - T)} = kt + C$


$T_{atm} - T = e^{(-kt - c)} = Ae^{-kt}$ (where A is a constant, defined by $A = e^{-c}$)


$T(t) = T_{atm} - Ae^{-kt}$

As we can see, the temperature $T(t)$ is not a linear function of time. This means that if the object is initially at 10 degrees and the atmosphere is at 30 degrees, then time taken to increase from 10 to 12 degrees will be different to the time taken to increase temperature from 12 to 14 degrees and so on.

But note that the time taken to go from 12 to 14 degrees depends upon the temperature of the object, and the temperature of the environment. That is, if you increase the temperature of the environment, you will decrease the time taken to go from 12 to 14 degrees, and you could theoretically decrease it enough so that it is the same as the time it took for temperature to change from 10 to 12 degrees in the previous scenario.

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You should state clearly that this law is empirical and approximately valid just for certain substance. – Siyuan Ren Feb 5 '13 at 12:59

In general, the answer will be no. A fundamental quantity to describe a physical system is energy. And the temperature flow $\delta T$ is tied to the energy flow $\delta Q$ via the equation $\delta T=\tfrac{1}{C(T)}\delta Q$. Now the the energy flow between two subsystems as well as the heat capacity $C(T)$ can essentially be aribtrarily complex.

The equations above come from thermodaynamics and to give a clearer answer one should employ a dynamical theory. A relevant relation to name here is Fouriers law which determines the heat flux in time through conductivity and the spatially resolved temperature gradient. In certain systems you can eliminate the energy quantity from the equations which leads to a differential equation expressing a self-consistency for the temperature evolution. By setting the left hand side of the heat equation to zero, you can work out the conditions for the temperature change to be constant. Here you want a constant temperature gradient. Everything works out well for the standard example ideal gas: The heat capactiy is constant and the energy expression simple. You speak of an "object", supposedly a solid material, so the heat capacity $C(T)$ might be have a wild dependency of temperature, making the situation difficult.

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Looking at this as a statistical mechanics problem, the partition function is shifted at higher temperature which can result in a change in heat capacity. Since the micro-states available depend on the structure of the material, accurately deriving the temperature dependent heat capacity can be quite nuanced.

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