1. **Stating the problem:** A metal bar initially at temperature $100$°F is placed in a room at constant temperature $30$°F. After $20$ minutes, the bar's temperature drops to $60$°F. We want to find the temperature $T$ of the bar at any time $t$ using Newton's law of cooling. 2. **Newton's law of cooling formula:** $$\frac{dT}{dt} = -k(T - T_0)$$ where: - $T$ is the temperature of the object at time $t$, - $T_0$ is the ambient temperature, - $k$ is a positive constant related to the cooling rate. 3. **Given values:** - Initial temperature: $T(0) = 100$ - Ambient temperature: $T_0 = 30$ - Temperature at $t=20$: $T(20) = 60$ 4. **Solving the differential equation:** Separate variables: $$\frac{dT}{T - T_0} = -k \, dt$$ Integrate both sides: $$\int \frac{1}{T - 30} dT = -k \int dt$$ $$\ln|T - 30| = -kt + C$$ Exponentiate: $$|T - 30| = e^{-kt + C} = Ae^{-kt}$$ where $A = e^C$. 5. **Apply initial condition $T(0) = 100$:** $$100 - 30 = A e^{0} = A$$ $$A = 70$$ So, $$T(t) = 30 + 70 e^{-kt}$$ 6. **Use $T(20) = 60$ to find $k$:** $$60 = 30 + 70 e^{-20k}$$ $$30 = 70 e^{-20k}$$ $$\frac{30}{70} = e^{-20k}$$ $$\frac{3}{7} = e^{-20k}$$ Take natural log: $$\ln\left(\frac{3}{7}\right) = -20k$$ $$k = -\frac{1}{20} \ln\left(\frac{3}{7}\right)$$ 7. **Final formula for temperature:** $$T(t) = 30 + 70 e^{-kt} = 30 + 70 e^{-\left(-\frac{1}{20} \ln\left(\frac{3}{7}\right)\right) t} = 30 + 70 \left(\frac{3}{7}\right)^{\frac{t}{20}}$$ **Answer:** The temperature of the bar at time $t$ minutes is $$\boxed{T(t) = 30 + 70 \left(\frac{3}{7}\right)^{\frac{t}{20}}}$$ This shows the temperature exponentially decays from $100$°F towards the ambient $30$°F over time.