During a football match, 120 g of sweat evaporates from a player's skin.

Step 1. (a) $120\ \text{g} = 0.12\ \text{kg}$, so $Q = mL = 0.12\ \text{kg} \times 2.4 \times 10^{6}\ \text{J/kg} = 2.88 \times 10^{5}\ \text{J}$.

Step 2. (b) $t = \dfrac{2.88 \times 10^{5}\ \text{J}}{100\ \text{J/s}} = 2880\ \text{s}$, and $\dfrac{2880\ \text{s}}{60\ \text{s/min}} = 48\ \text{min}$.