(Solved): According to Newton's law of cooling a cup of coffee cools according to this differential equation ...

According to Newton's law of cooling a cup of coffee cools according to this differential equation $\frac{\mathrm{d}}{\mathrm{d}t}{T}_{\text{coffee }}=k\left(T_{\text{coffee }}-T_{\text{ambient }}\right)$ where $T_{\text{coffee }}=$ temperature of coffee in Celsius as function of time $\mathbf{t}$ and $T_{\text{ambient }}=$ ambient temperature in Celsius fixed If a boiling cup of coffee at $95^{\circ }\mathrm{C}=T_{\text{coffee }}(0)$ was bought into the room with ambient temperature of $20^{\circ }\mathrm{C}$ then find time $t$ such that $T_{\text{coffee }}(\mathbf{t})=65^{\circ }\mathrm{C}$ given that at that time $t$ it is dropping at rate of $1^{\circ }\mathrm{C}$ per unit time