1. **State the problem:** We have coffee initially at $T_0 = 205^\circ$F placed in a room at ambient temperature $T_A = 72^\circ$F. After 10 minutes, the coffee temperature is $195^\circ$F. We want to find the time $t$ when the coffee cools to $180^\circ$F. 2. **Formula:** Newton's law of cooling is given by: $$T(t) = T_A + (T_0 - T_A)e^{-kt}$$ where $k$ is a positive constant. 3. **Find $k$ using the known data at $t=10$ minutes:** $$195 = 72 + (205 - 72)e^{-10k}$$ Simplify: $$195 - 72 = 133 e^{-10k}$$ $$123 = 133 e^{-10k}$$ Divide both sides by 133: $$\frac{123}{133} = e^{-10k}$$ Use \cancel to show division: $$\frac{\cancel{123}}{\cancel{133}} = e^{-10k}$$ Take natural logarithm on both sides: $$\ln\left(\frac{123}{133}\right) = -10k$$ Solve for $k$: $$k = -\frac{1}{10} \ln\left(\frac{123}{133}\right)$$ Calculate the value: $$k \approx -\frac{1}{10} \ln(0.9248) \approx -\frac{1}{10} (-0.0783) = 0.00783$$ 4. **Find time $t$ when $T(t) = 180^\circ$F:** $$180 = 72 + 133 e^{-kt}$$ $$180 - 72 = 133 e^{-kt}$$ $$108 = 133 e^{-kt}$$ Divide both sides by 133: $$\frac{108}{133} = e^{-kt}$$ Take natural logarithm: $$\ln\left(\frac{108}{133}\right) = -kt$$ Solve for $t$: $$t = -\frac{1}{k} \ln\left(\frac{108}{133}\right)$$ Substitute $k \approx 0.00783$: $$t \approx -\frac{1}{0.00783} \ln(0.8113)$$ Calculate: $$t \approx -127.7 \times (-0.209) = 26.7 \text{ minutes}$$ 5. **Answer:** Approximately 27 minutes, which corresponds to option C.