(Newton's Law of Cooling) Newton's law of cooling states that the rate of change of the temperature difference between an object and its surrounding medium is proportional to the temperature difference. If D(t) denotes this temperature difference at time t and if a denotes the constant of

Question

(Newton's Law of Cooling) Newton's law of cooling states that the rate of change of the temperature difference between an object and its surrounding medium is proportional to the temperature difference. If [latex]D(t)[/latex] denotes this temperature difference at time [latex]t[/latex] and if [latex]\alpha[/latex] denotes the constant of proportionality, then we obtain

[latex]\frac{d D}{d t}=-\alpha D ext {. }[/latex]

The minus sign indicates that this difference decreases. (If the object is cooler than the surrounding medium-usually air-it will warm up; if it is hotter, it will cool.) The solution to this differential equation is

[latex]D(t)=c e^{-\alpha t}[/latex].

If we denote the initial [latex](t=0)[/latex] temperature difference by [latex]D_{0}[/latex], then

[latex]D(t)=D_{0} e^{-\alpha t}[/latex]

is the formula for the temperature difference for any [latex]t>0[/latex]. Notice that for [latex]t[/latex] large [latex]e^{-\alpha t}[/latex] is very small, so that, as we have all observed, temperature differences tend to die out rather quickly.

We now may ask: In terms of the constant [latex]\alpha[/latex], how long does it take for the temperature difference to decrease to half its original value?

Accepted Answer

The original value is [latex]D_{0}[/latex]. We are therefore looking for a value of [latex]t[/latex] for which [latex]D(t)=\frac{1}{2} D_{0}[/latex]. That is, [latex]\frac{1}{2} D_{0}=D_{0} e^{-\alpha t}[/latex], or [latex]e^{-\alpha t}=\frac{1}{2}[/latex]. Taking natural logarithms, we obtain

[latex]-\alpha t=\ln \frac{1}{2}=-\ln 2 =-0.6931, \quad ext { and } \quad t \approx \frac{0.6931}{\alpha}[/latex].

Notice that this value of [latex]t[/latex] does not depend on the initial temperature difference [latex]D_{0}[/latex].

Question 11.2.4: (Newton's Law of Cooling) Newton's law of cooling states tha...

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