newton law of cooling

Question

A mug of coffee, with a temperature 𝑇℃ is made and left to cool in a room with a temperature of 25℃. The rate at which the coffee cools is proportional to the difference in temperature between the coffee and the room. Initially the coffee is at a temperature 85℃. 10 minutes later the coffee is at 55℃. Determine the temperature, to 1 decimal place, of the coffee after 15 minutes.

Answer 1

For T the instantaneous temperature of the coffee and t the time in minutes since the coffee started to cool, first write the rate of cooling as dT/dt equal to k(T − 25). For a Variables-Separable problem type, next write

dT/(T − 25) = kdt and integrate to ln |T − 25| = kt + ln |c|.

Then obtain ln |T − 25| − ln |c| equal to ln |(T − 25)/c| = kt. Rewrite as |(T − 25)/c| = e^kt or |T| = 25 + |c|e^kt.

For T = 85 at t = 0, write |85| = 25 + |c|e^(k×0) or |c| = 60, which gives T = 25 + 60e^kt. Note that T = 55 at t = 10

and state that 55 = 25 + 60e^(k×10) which simplifies to e^10k = 30/60 or 0.5.

Then 10k = ln 0.5 or k = (ln 0.5/10) and the instantaneous temperature of the coffee is given as

T = 25 + 60e^{(ln 0.5/10)t}.

At t = 15 minutes, obtain T equal to 25 + 60e^{(15ln 0.5/10)} or 46.21320344°C equivalent to 46.2°C.