1. **Problem statement:** A body initially at temperature 0F is placed in a room at 100F. After 10 minutes, the body's temperature is 25F. We need to find the time when the body's temperature reaches 50F. 2. **Formula used:** Newton's Law of Cooling states: $$\frac{dT}{dt} = k(T_m - T)$$ where $T$ is the temperature of the body at time $t$, $T_m$ is the ambient temperature, and $k$ is a constant. The solution to this differential equation is: $$T(t) = T_m - (T_m - T_0)e^{-kt}$$ where $T_0$ is the initial temperature. 3. **Given values:** - $T_0 = 0$ - $T_m = 100$ - $T(10) = 25$ 4. **Find $k$ using $T(10) = 25$:** $$25 = 100 - (100 - 0)e^{-10k}$$ $$25 = 100 - 100e^{-10k}$$ $$100e^{-10k} = 100 - 25 = 75$$ $$e^{-10k} = \frac{75}{100} = 0.75$$ Taking natural log: $$-10k = \ln(0.75)$$ $$k = -\frac{\ln(0.75)}{10}$$ 5. **Find time $t$ when $T(t) = 50$:** $$50 = 100 - 100e^{-kt}$$ $$100e^{-kt} = 100 - 50 = 50$$ $$e^{-kt} = 0.5$$ Taking natural log: $$-kt = \ln(0.5)$$ $$t = -\frac{\ln(0.5)}{k}$$ 6. **Substitute $k$ from step 4:** $$t = -\frac{\ln(0.5)}{-\frac{\ln(0.75)}{10}} = 10 \times \frac{\ln(0.5)}{\ln(0.75)}$$ 7. **Calculate numerical value:** $$\ln(0.5) \approx -0.6931$$ $$\ln(0.75) \approx -0.2877$$ $$t = 10 \times \frac{-0.6931}{-0.2877} = 10 \times 2.41 = 24.1$$ minutes **Final answer:** The body will reach 50F after approximately 24.1 minutes.