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

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?

• What do you mean by the constant of proportionality? In your equation, would that be "80.", or "-0.02179"? – David White Jun 20 '20 at 19:57

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:

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