Newton's Law of Cooling

Tags

Calculus

Cegep/2

Word count

304 words

Reading time

2 minutes

The cooling rate of an object is proportional to the difference between the temperature of the object and the ambient temperature.

$$\begin{array}{rl}\frac{\mathrm{d}T}{\mathrm{d}t}& \propto T-T_s\\ \frac{\mathrm{d}T}{\mathrm{d}t}& =k(T-T_s)\end{array}$$

Examples

A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is 350°𝐹. The temperature of the kitchen is 75°𝐹, and after 5 minutes the temperature of the pizza is 340°𝐹. We would like to wait until the temperature of the pizza reaches 300°𝐹 before cutting and serving it.

1. Set up the differential equation and the initial value problem modelling the temperature of the pizza.

We know that

$$\begin{array}{rl}{T}_s& =75\\ T(0)& =350\end{array}$$

We want to solve $\frac{\mathrm{d}T}{\mathrm{d}t}=k(T-75)$ given initial value $T(0)=350$.

2. Solve the differential equation to find the function 𝑇(𝑡).

$$\begin{array}{r}\\ \frac{\mathrm{d}T}{\mathrm{d}t}& =k(T-75)\\ \int \frac{1}{T-75}\mathrm{d}T& =\int k\mathrm{d}t\\ \ln (T-75)& =kt+C\\ T& =e^{kt+C}+75\end{array}$$

Substitute in $(0,350)$ to find $C$:

$$\begin{array}{rl}350& =e^{k\cdot 0+C}+75\\ C& =\ln 275\end{array}$$

Substitute in $(5,340)$ to find $k$:

$$\begin{array}{rl}340& =e^{5k+\ln 275}+75\\ 5k+\ln 275& =\ln 265\\ k& =\frac{\ln 265-\ln 275}{5}\\ k& =\frac{1}{5}\ln \frac{53}{55}\\ \\ \therefore T(t)& =e^{\frac{1}{5}\ln \frac{53}{55}t+\ln 275}+75\end{array}$$

3. How much longer will we need to wait before the pizza reaches a temperature of 300°𝐹?

$$\begin{array}{rl}300& =e^{\frac{1}{5}\ln \frac{53}{55}t+\ln 275}+75\\ \frac{1}{5}\ln \frac{53}{55}t+\ln 275& =\ln 225\\ t& =\frac{\ln 225-\ln 275}{\frac{1}{5}\ln \frac{53}{55}}\\ & =5\frac{\ln \frac{9}{11}}{\ln \frac{53}{55}}\\ & \approx 27.0874\min \end{array}$$

Contributors

Ajitani Hifumi

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