Given 3 points of heat of a thermometer and time, find the temperature of the oven?

Question

Just as a heads up: this is a homework question, so I would prefer guidance as opposed to a definitive answer.

The question poses that an oven is preheated to a constant temperature. A thermometer is placed inside starting at $26^\circ$C. After $40$ seconds the thermometer reads $32^\circ$C, and after $80$ it reads $33^\circ$C. How hot is the oven?

The class is differential equations, so I would presume it expects the use of the heat equation. I'm just not sure where to start. We also haven't started PDEs yet so I'm not sure how we are expected to answer this. It expects an answer to $2$ decimals.

Answer 1

You are presumably expected to assume the reading of the thermometer rises at a rate proportional to the temperature difference of the oven and the thermometer. Write the differential equation with an unknown constant of proportionality. Solve the equation, then use the data points to evaluate the constant. The result should exponentially approach the oven temperature.

Answer 2

Thanks to Bryce from my program and Ross Millikan's answer I was able to answer this question.

You must first build the equation. As the temperature change of the thermometer is proportional to the difference between the oven and the thermometer, we can write this as a constant multiplied by the temperature difference:

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

Where $t$ is the time in seconds, $T$ is the thermometer temperature, and $T_m$ is the oven temperature. We can rearrange this and integrate both sides to get:

$$\int\frac{1}{(T-T_m)}dT=\int kdt$$ $$ln|T-T_m|=kt+C$$

Then raise both sides as powers of $e$ and rearrange to get:

$$T=Ce^{kt}+T_m$$

Then we can start plugging in points. Using $(0, 26)$ yields $$26=Ce^{k(0)}+T_m$$ $$C=26-T_m$$

And then we can use this C along with the next point, $(40, 32)$: $$32=(26-T_m)e^{40k}+T_m$$ $$e^{40k}=\frac{32-T_m}{26-T_m}$$

Then do the same with the final point $(80, 33)$: $$33=(26-T_m)e^{80k}+T_m$$ $$e^{80k}=\frac{33-T_m}{26-T_m}$$

Notice that $e^{80k}=(e^{40k})^2$. We then use this fact to evaluate $T_m$: $$\frac{33-T_m}{26-T_m}=(\frac{32-T_m}{26-T_m})^2$$ $$(33-T_m)^2=(33-T_m)(26-T_m)$$ $$1024-64T_m+{T_m}^2=858-59T_m+{T_m}^2$$ $$T_m=33.2$$

This is your final answer. The temperature of the oven is $33.2^\circ$C.