# Temperature Problems

Newton's law of cooling, which is equally applicable to heating, states that the time rate of change of the temperature of a body is proportional to the temperature difference between the body and its surrounding medium. Let $$${T}$$$ denote the temperature of the body and $$${T}_{{m}}$$$ denote the temperature of the surrounding medium. Then, the time rate of change of the temperature of the body is $$$\frac{{{d}{T}}}{{{d}{t}}}$$$, and Newton's law of cooling can be formulated as $$$\frac{{{d}{T}}}{{{d}{t}}}=-{k}{\left({T}-{T}_{{m}}\right)}$$$, or as $$$\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}={k}{T}_{{m}}$$$, where $$${k}$$$ is a positive constant of proportionality. Once $$${k}$$$ is chosen positive, the minus sign is required in Newton's law to make $$$\frac{{{d}{T}}}{{{d}{t}}}$$$ negative in a cooling process, when $$${T}$$$ is greater than $$${T}_{{m}}$$$, and to make it positive in a heating process, when $$${T}$$$ is smaller than $$${T}_{{m}}$$$.

**Example 1.** A body at an unknown temperature is placed in a room which is held at a constant temperature of 30° F. If after 10 minutes the temperature of the body is 0° F and after 20 minutes the temperature of the body is 15° F, find the unknown initial temperature.

We have that $$${T}_{{m}}={30}$$$; so, the differential equation is $$$\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}={30}{k}$$$. This is a first-order linear differential equation. The integrating factor is $$${I}={{e}}^{{\int{k}{d}{t}}}={{e}}^{{{k}{t}}}$$$. After multiplying the equation by the integrating factor, we obtain that $$${{e}}^{{{k}{t}}}\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}{{e}}^{{{k}{t}}}={30}{k}{{e}}^{{{k}{t}}}$$$, or $$$\frac{{{d}{\left({T}{{e}}^{{{k}{t}}}\right)}}}{{{d}{t}}}={30}{k}{{e}}^{{{k}{t}}}$$$.

Integrating both sides gives $$${T}{{e}}^{{{k}{t}}}={30}{{e}}^{{{k}{t}}}+{C}$$$, or $$${T}={C}{{e}}^{{-{k}{t}}}+{30}$$$.

We are given that $$${T}{\left({10}\right)}={0}$$$, or $$${0}={C}{{e}}^{{-{10}{k}}}+{30}$$$. Also, $$${T}{\left({20}\right)}={15}$$$, or $$${15}={C}{{e}}^{{-{20}{t}}}+{30}$$$.

Thus, we have a system of two equations:

$$${\left\{\begin{array}{c}{C}{{e}}^{{-{10}{k}}}=-{30}\\{C}{{e}}^{{-{20}{k}}}=-{15}\\ \end{array}\right.}$$$

Dividing the first equation by the second gives $$${{e}}^{{{10}{k}}}={2}$$$. Now, from the first equation, we have that $$${C}=-{30}{{e}}^{{{10}{k}}}=-{30}\cdot{2}=-{60}$$$.

Finally, $$${T}_{{0}}={T}{\left({0}\right)}={C}{{e}}^{{-{k}\cdot{0}}}+{30}={C}+{30}=-{60}+{30}=-{30}$$$.

Let's take a look at another interesting example.

**Example 2.** A body at a temperature of 50° F is placed in an oven whose temperature is kept at 150° F. If after 10 minutes the temperature of the body is 75° F, find the time required for the body to reach a temperature of 100° F.

We have that $$${T}{\left({0}\right)}={50}$$$, $$${T}{\left({10}\right)}={75}$$$, $$${T}_{{m}}={150}$$$.

So, the differential equation is $$$\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}={150}{k}$$$. Again, as in example 1, this is a linear first-order differential equation. Its solution is $$${T}={C}{{e}}^{{-{k}{t}}}+{150}$$$.

Since $$${T}{\left({0}\right)}={50}$$$, we have that $$${50}={C}{{e}}^{{-{k}\cdot{0}}}+{150}$$$, or $$${C}=-{100}$$$.

Now, the equation has the form $$${T}=-{100}{{e}}^{{-{k}{t}}}+{150}$$$.

Since $$${T}{\left({10}\right)}={75}$$$, we have that $$${75}=-{100}{{e}}^{{-{10}{k}}}+{150}$$$, or $$${k}=-\frac{{{\ln{{\left({0.75}\right)}}}}}{{10}}$$$.

Finally, the equation has the following form: $$${T}=-{100}{{e}}^{{\frac{{\ln{{\left({0.75}\right)}}}}{{10}}{t}}}+{150}$$$.

Now, we need to find such $$${t}$$$ that $$${T}{\left({t}\right)}={100}$$$:

$$${100}=-{100}{{e}}^{{\frac{{{\ln{{\left({0.75}\right)}}}}}{{10}}{t}}}+{150}$$$, or $$${t}={10}\frac{{\ln{{\left({0.5}\right)}}}}{{{\ln{{\left({0.75}\right)}}}}}\approx{24.0942}$$$ minutes.