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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{T} denote the temperature of the body and Tm{T}_{{m}} denote the temperature of the surrounding medium. Then, the time rate of change of the temperature of the body is dTdt\frac{{{d}{T}}}{{{d}{t}}}, and Newton's law of cooling can be formulated as dTdt=−k(T−Tm)\frac{{{d}{T}}}{{{d}{t}}}=-{k}{\left({T}-{T}_{{m}}\right)}, or as dTdt+kT=kTm\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}={k}{T}_{{m}}, where k{k} is a positive constant of proportionality. Once k{k} is chosen positive, the minus sign is required in Newton's law to make dTdt\frac{{{d}{T}}}{{{d}{t}}} negative in a cooling process, when T{T} is greater than Tm{T}_{{m}}, and to make it positive in a heating process, when T{T} is smaller than Tm{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 Tm=30{T}_{{m}}={30}; so, the differential equation is dTdt+kT=30k\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}={30}{k}. This is a first-order linear differential equation. The integrating factor is I=e∫kdt=ekt{I}={{e}}^{{\int{k}{d}{t}}}={{e}}^{{{k}{t}}}. After multiplying the equation by the integrating factor, we obtain that ektdTdt+kTekt=30kekt{{e}}^{{{k}{t}}}\frac{{{d}{T}}}{{{d}{t}}}+{k}{T}{{e}}^{{{k}{t}}}={30}{k}{{e}}^{{{k}{t}}}, or d(Tekt)dt=30kekt\frac{{{d}{\left({T}{{e}}^{{{k}{t}}}\right)}}}{{{d}{t}}}={30}{k}{{e}}^{{{k}{t}}}.

Integrating both sides gives Tekt=30ekt+C{T}{{e}}^{{{k}{t}}}={30}{{e}}^{{{k}{t}}}+{C}, or T=Ce−kt+30{T}={C}{{e}}^{{-{k}{t}}}+{30}.

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

Thus, we have a system of two equations:

{Ce−10k=−30Ce−20k=−15{\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 e10k=2{{e}}^{{{10}{k}}}={2}. Now, from the first equation, we have that C=−30e10k=−30⋅2=−60{C}=-{30}{{e}}^{{{10}{k}}}=-{30}\cdot{2}=-{60}.

Finally, T0=T(0)=Ce−k⋅0+30=C+30=−60+30=−30{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(0)=50{T}{\left({0}\right)}={50}, T(10)=75{T}{\left({10}\right)}={75}, Tm=150{T}_{{m}}={150}.

So, the differential equation is dTdt+kT=150k\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=Ce−kt+150{T}={C}{{e}}^{{-{k}{t}}}+{150}.

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

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

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

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

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

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