Cooling by evaporation

During an hour of football, a player's body makes about $500\ \text{W}$ of heat. You will make the program find the heat that different amounts of sweat carry away, the sweat needed to carry away a whole hour's heat, and how much the body would warm in that hour with no sweating at all.

The idea. Each kilogram of sweat that evaporates takes $2.4 \times 10^{6}\ \text{J}$ from the skin: $Q = mL$. In one hour a $500\ \text{W}$ body makes $500\ \text{J/s} \times 3600\ \text{s}$. A $60\ \text{kg}$ body warms by $1\ ^\circ\text{C}$ for every $60 \times 3500\ \text{J}$ it keeps.

Your tasks.

1. Inside the loop, work out q, the heat in joules taken by grams of sweat.
2. Find sweat_needed, the grams of sweat that carry away one hour's heat at $500\ \text{W}$.
3. Find rise, how many degrees a $60\ \text{kg}$ body would warm in that hour with no sweating.