Question

The temperature of a cup of coffee t minutes after it has been poured is described by the function

C(t) = 20 + 60(0.95)^t .
a. How many minutes does it take for the coffee to cool to a temperature of d degrees?
b. About how fast is the temperature dropping when t = 1?

Answer 1

a) You want to find t such that
.. C(t) = d
.. d = 20 +60*0.95^t
.. (d -20)/60 = 0.95^t
.. log((d -20)/60) = t*log(0.95)
.. t = log((d -20)/60)/log(0.95) . . . . . . time to cool to d degrees (d > 20)

b) C'(t) = 60*0.95^t*ln(0.95)
.. C'(1) = 60*0.95*ln(0.95) ≈ -2.924 °C/min