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

The Problem Description

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

This is Temperature Profile equation for the given condition

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

The Final 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?

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

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

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