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
