1. **State the problem:** We have a heater with power rating 420 joules/second (watts) heating 20 grams of water from 20°C to 80°C. We need to find the time taken. 2. **Formula used:** The heat energy required to raise the temperature of a substance is given by: $$Q = mc\Delta T$$ where: - $m$ is the mass of the water, - $c$ is the specific heat capacity of water, - $\Delta T$ is the change in temperature. 3. **Important values and rules:** - Mass $m = 20$ grams = 0.02 kg (since 1000 grams = 1 kg) - Specific heat capacity of water $c = 4200$ J/kg°C - Temperature change $\Delta T = 80 - 20 = 60$°C - Power $P = 420$ J/s 4. **Calculate heat energy required:** $$Q = 0.02 \times 4200 \times 60 = 5040 \text{ joules}$$ 5. **Calculate time:** Power is energy per unit time, so time $t$ is: $$t = \frac{Q}{P} = \frac{5040}{420} = 12 \text{ seconds}$$ **Final answer:** The time taken to heat the water is **12 seconds**.