A 1200 W stove passes 75% of its heat to a pan of 2.0 kg of water at 25 °C.

Step 1. To the boil. $\dfrac{2.0\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times 75\ \text{K}}{900\ \text{W}} = 698\ \text{s}$, about $11.6$ minutes.

Step 2. Boil away $0.50\ \text{kg}$. $\dfrac{0.50\ \text{kg} \times 2.26 \times 10^{6}\ \text{J/kg}}{900\ \text{W}} = 1256\ \text{s}$, about $21$ minutes.