A 60.0 kg runner expends 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. Determine the volume of bodily fluid( assume it'swater) lost per hour. (at 37.0 Celcius, the latent heatof evaporation of water is 2.41 \times 10^{6} J/kg).

A 60.0 kg runner expends 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. Determine the volume of bodily fluid( assume it'swater) lost per hour. (at 37.0 Celcius, the latent heatof evaporation of water is 2.41 \times 10^{6} J/kg).

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asked 2021-05-20
A 60.0 kg runner expends 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. Determine the volume of bodily fluid( assume it'swater) lost per hour. (at 37.0 Celcius, the latent heatof evaporation of water is \(\displaystyle{2.41}\times{10}^{{{6}}}\) J/kg).

Answers (1)

2021-05-22
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 \(\displaystyle{2.41}\times{10}^{{{6}}}\frac{{J}}{{k}}{g}\)
\(\displaystyle{V}=\frac{{m}}{{r}_{{{w}}}}\)
\(\displaystyle={403}{c}{m}^{{{3}}}\).
0

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