1. **State the problem:** A thermometer initially reads 18°C inside a house. When placed outside where the air temperature is 30°C, after 3 minutes the thermometer reads 21°C. We need to find the thermometer reading after 6 minutes. 2. **Model the situation:** This is a Newton's Law of Cooling/Heating problem. The temperature $T(t)$ of the thermometer at time $t$ satisfies: $$\frac{dT}{dt} = k (T_a - T)$$ where $T_a = 30$°C is the ambient temperature, and $k$ is a constant. 3. **General solution:** The solution to this differential equation is: $$T(t) = T_a - (T_a - T_0) e^{-kt}$$ where $T_0 = 18$°C is the initial temperature. 4. **Use given data to find $k$:** At $t=3$ minutes, $T(3) = 21$°C. Substitute: $$21 = 30 - (30 - 18) e^{-3k}$$ $$21 = 30 - 12 e^{-3k}$$ Rearranged: $$12 e^{-3k} = 30 - 21 = 9$$ $$e^{-3k} = \frac{9}{12} = 0.75$$ Take natural log: $$-3k = \ln(0.75)$$ $$k = -\frac{1}{3} \ln(0.75)$$ 5. **Find temperature at $t=6$ minutes:** $$T(6) = 30 - 12 e^{-6k}$$ Since $e^{-3k} = 0.75$, then: $$e^{-6k} = (e^{-3k})^2 = 0.75^2 = 0.5625$$ So: $$T(6) = 30 - 12 \times 0.5625 = 30 - 6.75 = 23.25$$ **Final answer:** The thermometer reading after 6 minutes is **23.25°C**. This corresponds to option c.