1 Heating an Office Building (Newton’s Law of Cooling)

Background

Suppose that in Winter the daytime temperature in a certain office building is maintained at $70^{\circ}F$ . The heating is shut off at 10 P.M. and turned on again at 6 A.M. On a certain day the temperature inside the building at 2 A.M was found to be $65^{\circ}F$ . The outside temperature was $50^{\circ}F$ at 10 P.M. and had dropped to $40^{\circ}F$ by 6 A.M.

Problem

What was the temperature inside the building when the heat was turned on at 6 A.M.

Physical Information

Experiments show that the time rate of change of the temperature $T$ of a body $B$ (which conducts heat well, as, for example, a copper ball does) is proportional to the difference between $T$ and the temperature of the surrounding medium (Newton’s law of cooling).

Solution Step 1: Setting up a model

Let $T(t)$ be the temperature inside the building and $T_{A}$ the outside temperature (assumed to be constant in Newton’s Law). Then by Newton’s law,

\frac{dT}{dt}=k(T-T_{A})

(1)

Such experimental laws are derived under idealized assumptions that rarely hold exactly. However, even if a model seems to fit the reality only poorly (as in the present case), it will still give valuable qualitative information. To see how good a model is, the engineer will collect experimental data and compare them with the calculations from the model.

Solution Step 2: General Solution

We cannot solve (1) because we do not know $T_{A}$ , just that it varied between $50^{\circ}F$ and $40^{\circ}F$ , so we follow the Golden Rule: If you cannot solve your problem, try to solve a simpler one.We solve (1) with the unknown function $T_{A}$ replaced by the average of the two known values, or $45^{\circ}F$ . For physical reasons we may expect that this will give us a reasonable approximate value of $T$ at 6 A.M.

For constant $T_{A}=45$ (or any other constantvalue) the ODE (1) is separable. Separation, integration, and taking exponents gives the general solution

\frac{dT}{T-45}=k\,dt,\qquad\ln|T-45|=kt+c^{\ast},\qquad T(t)=45+ce^{kt}\qquad% (c=e^{c^{\ast}})

Solution Step 3: Particular Solution

We choose 10 P.M. to be $t=0$ . Then the given initial condition is $T(0)=70$ and yields a particular solution, call it $T_{p}$ . By substitution

T(0)=45+ce^{0}=70,\qquad c=70-45=25,\qquad T_{p}(t)=45+25e^{kt}.

Solution Step 4: Determination of $k$

We use $T(4)=65$ , where $t=4$ is 2 A.M. Solving algebraically for $k$ and inserting $k$ into $T_{p}(t)$ gives (see Figure LABEL:fig:temp)

T_{p}(4)=45+25e^{4k}=65,\qquad e^{4k}=0.8,\qquad k=\frac{1}{4}\ln 0.8=-0.056,% \qquad T_{P}(t)=45+25e^{-0.056t}.

Figure 1: Particular Solution (temperature)

Solution Step 4: Answer and Interpretation

6 A.M. is $t=8$ (namely 8 hours after 10 P.M.), and

T_{p}(8)=45+25e^{-0.056\cdot 8}=61.

Hence the temperature in the building dropped by $9^{\circ}F$ from $70^{\circ}F$ to $61^{\circ}F$ , a result that looks reasonable.