Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 5A / b / ii

Table of contents
- Solution ii

Solution

At the first half-temperature time, the solution of Newton's law of cooling gives

1 + (α - 1) e^{- k t_{1}} = \frac{α}{2}, t_{1} = \frac{1}{k} lo g \frac{2 ( α - 1 )}{α - 2} .

Since

α > 2

, this time is positive and finite. The instantaneous temperature increase supplies the new initial condition

T (t_{1} +) = α T_{0} /2 + β

. Restart the same first-order linear differential equation at

t_{1}

:

T (t) = T_{0} + [(\frac{α}{2} - 1) T_{0} + β] e^{- k (t - t_{1})}, t > t_{1} .

This is Newton cooling with an instantaneous temperature jump. The time translation in the exponential decay is essential: its prefactor is the excess immediately after the jump.

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