# 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.