Newton's Law of Cooling - Help me understand the constant of proportionality

Question

I was doing problems on the application of differential equations. This question popped out there.

It is a simple Separable Equation with initial and end condition given. I found out the temperature profile and plotted it.

Then, when the ambient temperature is changed, obviously the constant of integration has to be adjusted suitably right? I did it with the initial condition.

Now my question is will the constant of proportionality(k) in Newton's law remain same as found with the previous ambient temperature? Why?

Assuming k remains constant, I proceed with the problem and found the solution

But I can't get the intuition behind it. What determines the value of k? I found out k value by using an information pertaining to a different ambient temperature in the first place.

If I tweak the ambient temperature and the k value remains same, then is it safe to assume that all the cooling follows the same exponential path?

Answer 1

Just look at the Law in greater detail:

$$\frac{\text{d}Q}{\text{d}t}=-hA[T(t)-T_e]$$

which is the heat flow the cooling object loses to the environment, in $\mathrm{W}$.

Now an infinitesimal heat loss $\text{d}Q$ can also be written as:

$$\text{d}Q=mc_p\text{d}T(t)$$

where $m$ is the mass of the cooling object and $c_p$ its specific heat capacity.

So we have:

$$\text{d}T(t)=-\frac{hA}{mc_p}[T(t)-T_e]\text{d}t$$

$$\Rightarrow \frac{\text{d}T(t)}{T(t)-T_e}=-\frac{hA}{mc_p}\text{d}t$$

Integrate between $[0,T_0]$ and $[t,T(t)]$: $$\ln\Big[\frac{T(t)-T_e}{T_0-T_e}\Big]=-\frac{hA}{mc_p}t$$ Engineering handbooks often cite:

$$\boxed{\frac{hA}{mc_p}=\frac{1}{\tau}}$$ where $\tau$ is the characteristic time and $\frac{t}{\tau}$ is a dimensionless group ($\Pi$).

Thus:

$$\frac{T(t)-T_e}{T_0-T_e}=\exp\Big(-\frac{t}{\tau}\Big)$$

Now my question is will the constant of proportionality ($k$) in Newton's law remain same as found with the previous ambient temperature? Why?

So it is obvious that 'in theory' at least the constant $\frac{1}{\tau}$ (what you call $k$) is independent of all temperatures.

In reality, $T(t)$ may have some small effect on $h$ and $c_p$.

Finally we can write:

$$\boxed{T(t)=T_e+({T_0-T_e})\exp\Big(-\frac{t}{\tau}\Big)}$$