1. **Problem statement:** We want to find the time required to heat 0.2 liters of water from 15°C to 100°C using a 2 kW electric kettle with 95% efficiency. 2. **Relevant formula:** The energy needed to heat water is given by $$Q = mc\Delta T$$ where: - $m$ is the mass of water in kilograms, - $c$ is the specific heat capacity of water ($4186 \frac{J}{kg\cdot °C}$), - $\Delta T$ is the temperature change in °C. 3. **Calculate mass:** Since 1 liter of water has a mass of 1 kg, 0.2 liters corresponds to $$m = 0.2\,kg$$. 4. **Calculate temperature change:** $$\Delta T = 100 - 15 = 85\,°C$$. 5. **Calculate energy required:** $$Q = 0.2 \times 4186 \times 85 = 71162\,J$$. 6. **Calculate effective power:** The kettle's rated power is 2000 W, but efficiency is 95%, so effective power used for heating is $$P = 2000 \times 0.95 = 1900\,W$$. 7. **Calculate time:** Power is energy per unit time, so time $$t = \frac{Q}{P} = \frac{71162}{1900} \approx 37.45\,seconds$$. **Final answer:** It will take approximately **37.45 seconds** to heat the water to 100°C.