A 1.08 kW heater is used to convert 100 g of water at 100 °C into steam. How long will it take?

The power of the heater is $P = 1.08 \text{ kW}$.
The mass of steam to be produced is $m = 100 \text{ g}$.
The water is already at $100 \text{ °C}$, so we only need to consider the latent heat of vaporization.
The latent heat of vaporization of water ($L_v$) is approximately $2.26 \times 10^6 \text{ J/kg}$.

The heat energy $Q$ required for vaporization is given by the formula $Q = m L_v$.

The power $P$ of the heater is the rate at which it supplies energy, so $P = \frac{Q}{t}$, where $t$ is the time.
Rearranging for $t$:

The calculated time is approximately $209.26 \text{ s}$, which is closest to $210 \text{ s}$.