A 60.0 kg runner expends( P = 300 W) of power whilerunning. 10% of the energy is delivered to the musclesand 90% of the energy is primarily removed from the body bysweating.

Here ;

Power (P ) = 300 W

Time ( t ) = 3600 s

The latent heat of evaporation of water ( \(\displaystyle{L}_{{{v}}}={2.41}\times{10}^{{{6}}}\) J/kg).

The energy dissipated by the person ( E ) = P *t

\(\displaystyle={1.08}\cdot{10}^{{{6}}}{J}\)

The energy primarily removed from the body bysweating

Q = 90 % ( \(\displaystyle{1.08}\cdot{10}^{{{6}}}{J}\))

\(\displaystyle={9.72}\cdot{10}^{{{5}}}{J}\)

The mass of bodily fluid( assume it's water) lost per hour. (at 37.0 Celcius),the latent heat ofevaporation of water is \(\displaystyle{2.41}\times{10}^{{{6}}}\frac{{J}}{{k}}{g}\).

m = Q / Lv

\(\displaystyle={9.72}\cdot{10}^{{{5}}}\frac{{J}}{{2.41}}\times{10}^{{{6}}}\frac{{J}}{{k}}{g}\)

= 0.403 kg

The volume of bodily fluid( assume it's water) lost per hour. (at 37.0 Celcius, the latent heat ofevaporation of water is \(\displaystyle{2.41}\times{10}^{{{6}}}\frac{{J}}{{k}}{g}\)

\(\displaystyle{V}=\frac{{m}}{{r}_{{{w}}}}\)

\(\displaystyle={403}{c}{m}^{{{3}}}\).

Here ;

Power (P ) = 300 W

Time ( t ) = 3600 s

The latent heat of evaporation of water ( \(\displaystyle{L}_{{{v}}}={2.41}\times{10}^{{{6}}}\) J/kg).

The energy dissipated by the person ( E ) = P *t

\(\displaystyle={1.08}\cdot{10}^{{{6}}}{J}\)

The energy primarily removed from the body bysweating

Q = 90 % ( \(\displaystyle{1.08}\cdot{10}^{{{6}}}{J}\))

\(\displaystyle={9.72}\cdot{10}^{{{5}}}{J}\)

The mass of bodily fluid( assume it's water) lost per hour. (at 37.0 Celcius),the latent heat ofevaporation of water is \(\displaystyle{2.41}\times{10}^{{{6}}}\frac{{J}}{{k}}{g}\).

m = Q / Lv

\(\displaystyle={9.72}\cdot{10}^{{{5}}}\frac{{J}}{{2.41}}\times{10}^{{{6}}}\frac{{J}}{{k}}{g}\)

= 0.403 kg

The volume of bodily fluid( assume it's water) lost per hour. (at 37.0 Celcius, the latent heat ofevaporation of water is \(\displaystyle{2.41}\times{10}^{{{6}}}\frac{{J}}{{k}}{g}\)

\(\displaystyle{V}=\frac{{m}}{{r}_{{{w}}}}\)

\(\displaystyle={403}{c}{m}^{{{3}}}\).