Estabon poured himself a hot beverage that had a temperature of 198 then set it on the kitchen table to cool. The temperature of the kitchen was a constant 68. Of the drink cooled to 182 in five minutes how long will it take for the drink to cool to 90?

Simple sol'n:

Rate of decrease in temperature: - 16/5 °C/min

182 + x (-3.2) = 90

-3.2x = - 92

x = 28.75 min's

Total time to cool to 90 °C = 28.75 + 5 = 33.75 min's

By Differential Equations:

dT/dt = - kT, t = time, T = temperature of beverage

If you don't understand what I've written above, it is simply:

'The change in T with respect to t is proportional to T'

or 'The rate of cooling is proportional to T'

Continuing:

Separating the variables:

1/T dT = - k dt

Integrate both sides:

∫1/T dT = ∫-k dt

ln (T) = - kt + c

From what we're given: when t = 0, T = 198

ln (198) = - k (0) + c

c = ln (198)

so ...

ln (T) = - kt + ln (198)

Finding k:

ln (182) = - k (5) + ln (198)

-5k = ln (182/198) [ln (a) - ln (b) = ln (a/b) ]

k = - (ln (182/198)) / 5 = 0.01685 ...

Finding desired t value:

ln (T) = - kt + ln (198)

ln (90) = - (0.01685 ...) t + ln (198)

-0.01685 ... t = ln (90/198)

t = ln (5/11) / - 0.01685 ...

t = 46.79 min's

Total time = 46.79 + 5 = 51.79 min's