1. **State the problem:** A cup of tea cools from 90°C, losing 10% of its temperature difference from room temperature each minute. The room temperature is 20°C. We need to find the temperature of the tea after 6 minutes. 2. **Formula and explanation:** The cooling follows Newton's Law of Cooling, where the temperature difference decreases by a fixed percentage each minute. If $T_n$ is the temperature after $n$ minutes, and $T_r$ is the room temperature, then: $$T_n = T_r + (T_0 - T_r) \times (1 - r)^n$$ where: - $T_0 = 90$°C (initial temperature) - $T_r = 20$°C (room temperature) - $r = 0.10$ (10% loss per minute) - $n = 6$ (minutes) 3. **Calculate the temperature difference initially:** $$\Delta T_0 = T_0 - T_r = 90 - 20 = 70$$ 4. **Calculate the temperature difference after 6 minutes:** $$\Delta T_6 = 70 \times (1 - 0.10)^6 = 70 \times 0.9^6$$ Calculate $0.9^6$: $$0.9^6 = 0.531441$$ So, $$\Delta T_6 = 70 \times 0.531441 = 37.20087$$ 5. **Find the temperature after 6 minutes:** $$T_6 = T_r + \Delta T_6 = 20 + 37.20087 = 57.20087$$ 6. **Final answer:** The temperature of the tea after 6 minutes is approximately **57.2°C**.