#### Answer

(a) The temperature after 50 minutes is $13.3^{\circ}C$
(b) The temperature will be $15^{\circ}C$ after 67.7 minutes.

#### Work Step by Step

(a) We can find $k$:
$T(t) = 20-15~e^{kt}$
$T(25) = 20-15~e^{25k} = 10$
$-15~e^{25k} = -10$
$e^{25k} = \frac{10}{15}$
$25k = ln(\frac{2}{3})$
$k = \frac{ln(\frac{2}{3})}{25}$
$k = -0.0162186$
Then:
$T(t) = 20-15~e^{-0.0162186~t}$
We can find the temperature after 50 minutes:
$T(t) = 20-15~e^{-0.0162186~t}$
$T(50) = 20-15~e^{(-0.0162186)~(50)}$
$T(50) = 13.3$
The temperature after 50 minutes is $13.3^{\circ}C$
(b) We can find the time $t$ when the temperature is $15^{\circ}C$:
$T(t) = 20-15~e^{-0.0162186~t} = 15$
$e^{-0.0162186~t} = \frac{1}{3}$
$-0.0162186~t = ln(\frac{1}{3})$
$t = \frac{ln(\frac{1}{3})}{-0.0162186}$
$t = 67.7~minutes$
The temperature will be $15^{\circ}C$ after 67.7 minutes.