I am trying to prove the Newton’s Law of cooling equation -

$$\frac{dT}{dt} = -k(T - T_a)$$ where $T(t)$ is the temperature of the object at time $t$, $T_a$ is the ambient temperature, and $k$ is a positive constant.

I did a simple home experiment by boiling water and letting it cool in room temperature (ambient temperature - 25 degrees Celsius)

I ended up with an exponential decay data/graph and to prove that, I have done an integration factor method to integrate the Newton’s Law of cooling function.

Then I was wondering, will there be any assumptions that I have to make?

For example, do I assume that there is no convective heat exchange but rather only conductive heat exchange due to the temperature difference?

In addition, do I take into consideration the type of material of the table that I put the beaker on? Or to I assume it to have no effect in other to satisfy the Newton law of cooling?

  • 1
    $\begingroup$ The Wiki article is pretty clear on the matter. Newton's law of cooling assumes a constant heat transfer coefficient. This is generally true for conduction, often an okay approximation for convection, and false for thermal radiation. $\endgroup$
    – lemon
    Jun 30, 2018 at 8:27

1 Answer 1


Newton's law of cooling assumes that the temperature variations within the system (in this case the fluid in the beaker) are negligible compared to the temperature difference between the system and its surroundings. The temperature variations within the system will be negligible if there is significant natural or forced convection (stirring) within this system and/or high thermal conductivity of the material comprising the system. In this case, the main resistance to heat transfer will be between the surface of the system and the surroundings. The constant k in Newton's law is given by: $$k=\frac{\int{hdA}}{\rho C_p}$$where h is the local heat transfer coefficient between the system and surroundings over each differential interface area dA, $\rho$ is the liquid density, and $C_p$ is the liquid thermal conductivity. Assuming that the table is at the same temperature as the surrounding air, h in the region of contact with the table will be higher than h in the region directly in contact with the surrounding air. Note that h also accounts for the thermal resistance of the glass comprising the wall of the beaker.


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