Math 2.0

example

Newton's law of cooling

Example

Let T(t)T(t) be the temperature of a body at time tt. Let TaT_a be the constant temperature of the surroundings, and let k>0k > 0 be a constant. Newton's law of cooling is the equation T′(t)=−k(T(t)−Ta).T'(t) = -k \left(T(t) - T_a\right). With the initial condition T(0)=T0T(0) = T_0, its unique solution is T(t)=Ta+(T0−Ta) e−kt.T(t) = T_a + (T_0 - T_a)\, e^{-k t}.

Remarks

The difference D=T−TaD = T - T_a satisfies D′=−kDD' = -k D. This is exponential decay, so D(t)=D(0)e−ktD(t) = D(0) e^{-kt}. As t→∞t \to \infty, T(t)→TaT(t) \to T_a. If T0>TaT_0 > T_a the body cools, and if T0<TaT_0 < T_a it warms. The law is a good model when the temperature difference is small and the body has a nearly uniform temperature.

Explanations

L1 human

A hot cup of coffee cools fast at first and then more slowly. The law says that the speed of cooling is proportional to the difference between the coffee and the room. So the difference halves after each fixed time interval.

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Sources

  1. Web https://en.wikipedia.org/wiki/Newton%27s_law_of_cooling, First-order transient response of lumped-capacitance objects

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