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

When rates are expressed as a quantity of 1, such as 2 feet per second or 5 miles per hour, they are called unit rates. if you have a multiple-unit rate such as 120 students for every 3 buses, and want to find the single-unit rate, write a ratio equal to the multiple-unit rate with 1 as the second term.