8. A jogger generates heat energy at a rate of *735.3* W. If all this energy is removed by sweating, how much water must evaporate from the jogger's skin each hour?

The heat energy generated by the jogger is *735.3* W and this energy is used to evaporate water from the skin.

The amount of energy required to evaporate water can be calculated using the formula:

Q = mL

where:
Q = heat energy (in J),
m = mass of water (in kg) evaporated,
L = specific latent heat of vaporization of water = 2260 J/g

Since the jogger generates 735.3 W of heat energy in 1 hour, we can convert this to Joules:

735.3 \, \text{W} \times 1 \, \text{hour} \times 3600 \, \text{s/hour} = 2647080 \, \text{J}

Now, we can find the mass of water evaporated using the formula:

m = \frac{Q}{L}

m = \frac{2647080 \, \text{J}}{2260 \, \text{J/g}}

m = 1172.21 \, \text{g} = 1.17221 \, \text{kg}

Therefore, the jogger must evaporate approximately 1.17221 kg of water per hour to remove the generated heat energy.

\boxed{1.17221 \, \text{kg}} of water must evaporate from the jogger's skin each hour.