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.

  • $\begingroup$ 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. $\endgroup$ – dshapiro Mar 24 '15 at 14:33
  • $\begingroup$ One hint is that the value of k remains the same. Equate those values to get T0. $\endgroup$ – 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.

  • $\begingroup$ What does the answer come out to? $\endgroup$ – 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


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


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









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


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