Hot water in a calorimeter cools in a room at 25 °C…

Step 1. The cooling constant. The excess falls from $45$ K to $33$ K in $10$ min. $k = \dfrac{1}{10\ \text{min}}\ln\dfrac{45}{33}$ $= 0.0310\ \text{min}^{-1}$.

Step 2. The half-time. $t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{0.0310\ \text{min}^{-1}}$ $= 22.3$ min.

Step 3. After 20 minutes. $\theta = 25 + 45\,e^{-0.0310 \times 20}$ °C $= 25 + 45 \times 0.538 = 49.2$ °C.

Step 4. To reach 40 °C. The excess is $15$ K, one third of $45$ K. So $t = \dfrac{\ln 3}{k} = \dfrac{1.099}{0.0310\ \text{min}^{-1}}$ $= 35.4$ min.