Answer
(a) $46.7^{\circ}C$
(b) The hot chocolate will have cooled to $40^{\circ}C$ after 81.3 minutes.
Work Step by Step
(a) We can find $k$:
$T(t) = 20+60~e^{kt}$
$T(30) = 20+60~e^{30k} = 60$
$60~e^{30k} = 40$
$e^{30k} = \frac{40}{60}$
$30k = ln(\frac{2}{3})$
$k = \frac{ln(\frac{2}{3})}{30}$
$k = -0.0135155$
Then:
$T(t) = 20+60~e^{-0.0135155~t}$
We can find the temperature after a total of 60 minutes:
$T(t) = 20+60~e^{-0.0135155~t}$
$T(60) = 20+60~e^{(-0.0135155)(60)}$
$T(60) = 46.7$
The temperature after 60 minutes is $46.7^{\circ}C$
(b) We can find the time $t$ when the temperature is $40^{\circ}C$:
$T(t) = 20+60~e^{-0.0135155~t}$
$20+60~e^{-0.0135155~t} = 40$
$60~e^{-0.0135155~t} = 20$
$e^{-0.0135155~t} = \frac{20}{60}$
$-0.0135155~t = ln(\frac{1}{3})$
$t = \frac{ln(\frac{1}{3})}{-0.0135155}$
$t = 81.3$
The hot chocolate will have cooled to $40^{\circ}C$ after 81.3 minutes.
