# Differential equations: Newton's Law of Coolin

Submitted by Anisul Mahmud on April 18, 2021 - 4:19pm

Need help on this..DE Newton's Law of Cooling

A thermometer is removed from a room whose ambient temperature is 70° F and is taken outside.

After one minute the thermometer reads 60° F and after three minutes, it reads 55° F. **What is the outside
temperature (which is assumed to be constant)**? Find your answer by solving DE model of Newton’s Law

of cooling.

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## You may find this useful:

You may find this useful: https://mathalino.com/node/2284

## Hello!

Hello!

To get the temperature outside when the thermometer reading is dropping from 70 $^{o}$ F to 60 $^{o}$ F to 55 $^{o}$ F within three minutes, we need to use the Newton's Law of Cooling, but this isn't the only way to get the outside temperature.

First, we need to get the expression of the final temperature so we can get the outside temperature. We already know the final temperatures (70 $^{o}$ F at 0 minutes, 60 $^{o}$ F at 1 minutes and 55 $^{o}$ F at 3 minutes), so we can get the outside temperature too.

The temperature of many objects can be modelled using a differential equation. Newton's law of cooling (or heating) states that the temperature of a body changes at a rate proportional to the difference in temperature between the body and its surroundings. It is a reasonably accurate approximation in some circumstances.

Let $T$ denote the temperature of an object and $T_{o}$ the ambient temperature. If $t$ denotes time, then Newton's law states that...

$$ \frac {Dt}{dt} = -k(T - T_o)$$

Rearranging the equation as

$$Dt = -k(T - T_o) dt$$

..and simplifying it a bit....it becomes...

$$\frac{dT}{T-T_o} = -k dt$$

Now doing some implicit differentiation to get the expression of final temperature...

$$\int \frac{dT}{T-T_o} = \int -k dt$$

$$ln(T - T_o) = -kt + C$$

$$ln(T - T_o) = ln e^{-kt} + lnC$$

$$ln(T - T_o) = ln Ce^{-kt}$$

$$(T - T_o) = Ce^{-kt}$$

$$T = T_o + Ce^{-kt}$$

The expression of the final temperature, which is the solution for the differential equation of Newton's Law of Cooling is....

$$T = T_o + Ce^{-kt}$$

Now we got the final temperature, we can try chugging in the given information and see if we could come up with an answer. Those information were the final temperatures 70 $^{o}$ F at 0 minutes, 60 $^{o}$ F at 1 minutes and 55 $^{o}$ F at 3 minutes. With that in mind, we can now chug them in...

When the thermometer was put outside, at that instant, the temperature is 70 $^{o}$ F. So...at $t = 0$, the temperature of thermometer is 70 $^{o}$ F. The expression of the final temperature would be....

$$70 = T_o + Ce^{-k(0)}$$

$$70 = T_o + C$$

When the thermometer's temperature was measured after one minute, it was now 60 $^{o}$ F. So...at $t = 1$, the temperature of thermometer is 60 $^{o}$ F. The expression of the final temperature would be....

$$60 = T_o + Ce^{-k(1)}$$

$$60 = T_o + Ce^{-k}$$

When the thermometer's temperature was measured after three minutes, it was now only 55$^{o}$ F. So...at $t = 3$, the temperature of thermometer is 55 $^{o}$ F. The expression of the final temperature would be....

$$55= T_o + Ce^{-k(3)}$$

$$55 = T_o + Ce^{-3k}$$

At this point, it looks nice because we have three equations with three unknowns. The $T_o$, the $k$ and the $C$. The three equations were...

$$70 = T_o + C$$

$$60 = T_o + Ce^{-k}$$

$$55 = T_o + Ce^{-3k}$$

To get the three unknowns, we can solve them....One of the ways to get the unknowns look like this (and were really interested in getting the ambient temperature $T_o$)

Rearranging the first equation $70 = T_o + C$, it becomes...

$$70 - T_o = C$$

Rearranging the first second equation $60 = T_o + Ce^{-k}$, it becomes...

$$60 - T_o = Ce^{-k}$$

Rearranging the first third equation $55 = T_o + Ce^{-3k}$, it becomes...

$$55 - T_o = Ce^{-3k}$$

Chugging the expression $70 - T_o = C$ into the second equation $60 - T_o = Ce^{-k}$, it becomes....

$$60 - T_o = Ce^{-k}$$

$$60 - T_o = (70 - T_o)e^{-k}$$

Now getting the $e^{-k}$, it becomes....

$$e^{-k} = \frac{60 - T_o}{70 - T_o}$$

Notice that in the third equation $55 - T_o = Ce^{-3k}$, the expression $e^{-3k}$ is just a fancy way of writing as $(e^{-k})^3$. We now realize that we can write the third expression purely in terms of $T_o$'s. To show it, here it is....

$$55 - T_o = Ce^{-3k}$$

$$55 - T_o = C(e^{-k})^3$$

$$55 - T_o = (70 - T_o) \left( \frac{60 - T_o}{70 - T_o}\right )^3$$

$$55 - T_o = \frac{(60 - T_o)^3}{(70 - T_o)^2}$$

Now getting the $T_o$, which is the outside temperature we have been seeking by shift - solving in advanced calculators..is now...$\color{blue}{54.23 ^o F}$.

Alternate solutions are encouraged....:-)

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