All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

$\frac{d \theta}{dy} = k(20-\theta)$ where k is a positive constant.

When $\theta = 10, \frac{d\theta}{dt}=1$

$\Rightarrow k = 0.1$

$\int \frac{1}{20-\theta} d\theta = \int 0.1 dt$

$-\mathrm{ln}|20-\theta| = 0.1t + C$

Since $\theta \le 20, -\mathrm{ln}(20-\theta) = 0.1t + C$

When $t = 0, \theta = 10$

$\Rightarrow C = -\mathrm{ln}10$

$\mathrm{ln}(20-\theta) = -0.1t + ln10$

$\theta = 20 - 10 e^{-0.1t}$

When $\theta = 15, t = 6.93\mathrm{~min}$

Not a tough question, just be care when you draw the graph, consider only the positive t values. Students can omit the module too since $\theta \le 20$ always.