# Newton’s Law of Cooling/Warming

A thermometer is placed in an oven preheated to a constant temperature of 390◦ F. Through a glass window in the oven door, an observer records that the thermometer reads 190◦ F after 1 minute and 230◦ F after 2 minutes. What is the initial reading of the thermometer?

I know that you have to use the formula $\frac{dT}{dt}=k(T-T_m)$. I don't know what to do next.

• Do you know how to solve the differential equation to get temperature as a function of time? I'm trying to understand which part of this solution you need help with. – dshapiro Mar 24 '15 at 14:33
• One hint is that the value of k remains the same. Equate those values to get T0. – Krishna Mar 24 '15 at 14:39

Newton's law of cooling would lead you to the differential equation,

$$\frac{dT}{dt} = k(T-T_i)$$

This is a very simple differential equation, which could be solved as:

$$\frac{dT}{T-T_i} = k dt$$ $$\Rightarrow \int_0^{t'} \frac{dT}{T-T_i} = \int_0^{t'} k dt$$ $$\Rightarrow \ln\left(\frac{T}{T_i}\right) = k t'$$ $$\Rightarrow T = T_i \exp (k t')$$

Now, the thing with exponential functions is, any point for which you know the values of $(t, T(t))$ can be taken as a starting point. I leave the rest to you as a simple exercise.

• What does the answer come out to? – Jessie Mar 24 '15 at 16:57

Newton's law serves equally for cooling or warming situations: $\frac{dT}{dt}=k(T_e-T)$

The general solution is $T(t)=T_e+(T_0-T_e)e^{-kt}$; where $T_0=T(0)$
is the initial reading we are going to find

In this case $T_e=390^{\circ}F$ is the oven temperature; so we can write

$T(t)=390+(T_0-390)e^{-kt}$

Now, we have recorded two temperatures at some given moments:

$190=390+(T_0-390)e^{-k(1)}$
$230=390+(T_0-390)e^{-k(2)}$

We have to solve these two last equations to find $T_0$ and $k$

$(T_0-390)e^{-k(1)}=-200$
$(T_0-390)e^{-k(2)}=-160$

$e^{-k}=\frac{200}{390-T_0}$
$e^{-2k}=\frac{160}{390-T_0}$

$-k=ln\frac{200}{390-T_0}$
$-2k=ln\frac{160}{390-T_0}$

$ln\frac{160}{390-T_0}=2ln\frac{200}{390-T_0}$

$ln\frac{160}{390-T_0}=ln(\frac{200}{390-T_0})^2$

$\frac{160}{390-T_0}=(\frac{200}{390-T_0})^2$

$390-T_0=\frac{200^2}{160}=250$

$T_0=140^{\circ}F$

With this value we can find $k=ln(1.25)$