A 250 g sample of water at 25°C is to be boiled using an electric boiling ring with a voltage of 220 V and a resistance of 10 Ω. Calculate the time required to boil the water. Assume the specific heat

The water is heated from $25^\circ\text{C}$ to its boiling point, which is $100^\circ\text{C}$.

$$\Delta T = T_{\text{final}} - T_{\text{initial}} = 100^\circ\text{C} - 25^\circ\text{C} = 75^\circ\text{C}$$

Since $1^\circ\text{C}$ change is equal to $1\,\text{K}$ change, $\Delta T = 75\,\text{K}$.

The mass of water is $250\,\text{g}$, which is $0.250\,\text{kg}$. The specific heat capacity of water is $4000\,\text{J/kgK}$.

$$Q = mc\Delta T$$ $$Q = (0.250\,\text{kg})(4000\,\text{J/kgK})(75\,\text{K})$$ $$Q = 75000\,\text{J}$$

The voltage is $220\,\text{V}$ and the resistance is $10\,\Omega$.

$$P = \frac{V^2}{R}$$ $$P = \frac{(220\,\text{V})^2}{10\,\Omega}$$ $$P = \frac{48400\,\text{V}^2}{10\,\Omega}$$ $$P = 4840\,\text{W}$$

Assuming all electrical energy is converted to heat in the water, the electrical energy supplied ($E$) is equal to the heat energy required ($Q$).

$$E = P \times t$$ $$Q = P \times t$$ $$75000\,\text{J} = (4840\,\text{W}) \times t$$ $$t = \frac{75000\,\text{J}}{4840\,\text{W}}$$ $$t \approx 15.50\,\text{s}$$

The time for which the electric boiling ring should be operated is $\boxed{\text{15.50 s}}$.