The problem statement says that the heating element is on for a long time, so we can assume that the temperature of the water is constant. This means that the water loses $100 \mbox{ W}$ to the environment. When the heating element is removed, the rate that energy is lost will be approximately the same at least during the amount of time it takes the water to cool by $1 \;^{\circ}\mathrm{C}$ . The density of water is approximately $1000 \;\mathrm{kg}\mathrm{/}\mathrm{m}^3$ and one liter is $10^{-3} \;\mathrm{m}^3$ . Therefore, the mass of water in the pan is

$\begin{align*}1 L \cdot \frac{10^{-3} \;\mathrm{m}^3}{L} \cdot 1000 \;\mathrm{kg}\mathrm{/}\mathrm{m}^3 = 1 \mbox{ kg}\end{al...$
So, the answer is
$\begin{align*}\frac{1 \mbox{ kg} \cdot 1 \;^{\circ}\mathrm{C} \cdot 4.2 \cdot 10^3 \;\mathrm{J}\mathrm{/}\mathrm{(}\mathrm{kg...$
Therefore, answer (B) is correct.

Solution to 1996 Problem 13