1. **State the problem:** A cup of coffee initially at 95°C is cooling in a room at 20°C. When the coffee's temperature is 72°C, it is cooling at a rate of 1°C per minute. We need to find how many minutes have passed when the temperature is 72°C. 2. **Use Newton's Law of Cooling:** The rate of change of temperature is proportional to the difference between the object's temperature and the ambient temperature: $$\frac{dT}{dt} = -k(T - T_{room})$$ where $T$ is the temperature of the coffee at time $t$, $T_{room} = 20$, and $k$ is a positive constant. 3. **General solution:** $$T(t) = T_{room} + (T_0 - T_{room})e^{-kt}$$ where $T_0 = 95$ is the initial temperature. 4. **Find $k$ using the given cooling rate at $T=72$:** At $T=72$, the rate of cooling is $\frac{dT}{dt} = -1$ °C/min. From the differential equation: $$\frac{dT}{dt} = -k(T - 20)$$ Plug in $T=72$ and $\frac{dT}{dt} = -1$: $$-1 = -k(72 - 20) = -k \times 52$$ So, $$k = \frac{1}{52}$$ 5. **Find time $t$ when $T=72$:** Use the temperature formula: $$72 = 20 + (95 - 20)e^{-\frac{1}{52}t}$$ Simplify: $$72 - 20 = 75 e^{-\frac{1}{52}t}$$ $$52 = 75 e^{-\frac{1}{52}t}$$ Divide both sides by 75: $$\frac{52}{75} = e^{-\frac{1}{52}t}$$ Take natural logarithm: $$\ln\left(\frac{52}{75}\right) = -\frac{1}{52}t$$ Multiply both sides by $-52$: $$t = -52 \ln\left(\frac{52}{75}\right)$$ 6. **Calculate $t$:** $$t = -52 \times \ln(0.6933) \approx -52 \times (-0.3662) = 19.04$$ **Answer:** It takes approximately **19.04 minutes** for the coffee to cool to 72°C.