I performed the simple Newton cooling experiment where I have a body of known surface area $A$. If I heat it and then place it hanging from somewhere, so that it is in contact only with air, it will cool exponentially. Here I am able to measure the thermal conductivity of the material $h$, where $k=Ah/mc$ ($m$ the mass of the object and $c$ it's specific heat), and the temperature decreases as $e^{-kt}$. Now, from the construction of the Newton's model

$$A^{-1}\frac{dQ}{dt} = h(T-T_a), \quad \quad \quad \quad dQ=-mcdT$$

I never take into account the medium that absorbs the heat from the body. Of course $h$ is not the thermal conductivity of the material, since it depends on the material of the body, and also on the material of the medium. How can I extract information (i.e. some coefficients of thermal conductivity) from the medium and from the body separately?

I appreciate your help.


1 Answer 1


Usually, when Newton's law of cooling is used, it is assumed that the main resistance to heat transfer resides outside the body, and that the heat transfer coefficient h characterizes this outside resistance. This is not a bad approximation if the conductive resistance inside the body is small compared to the outside resistance. However, it is possible to also include the heat conduction inside the body, but, to do so, you need to solve the partial differential equation for transient heat transfer inside the body in conjunction with the convective boundary condition at the surface of the body. The solution to this analysis will not be an exponential decay, except at long times. So the nature of the solution at short times is different. It is however possible to get the asymptotic solution at long times, and use this to deduce the individual contributions of the inside- and outside resistances. But, if the outside resistance is much higher than the inside resistance, it may be difficult to resolve the inside resistance from the data. On the other hand, if that is the case, then the inside resistance is not very significant, and can be neglected. This takes you back to the Newton's cooling approximation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.