A 60.0-kg runner expends 335 W of power while running a marathon. Assuming that

Question

A 60.0-kg runner expends 335 W of power while running a marathon. Assuming that 9.00% of the energy is delivered to the muscle tissue and that the excess energy is removed from the body by sweating, determine the volume of bodily fluid (assume it is water) lost per hour. (At 37.0

A 60.0-kg runner expends 335 W of power while running a marathon. Assuming that 9.00% of the energy is delivered to the muscle tissue and that the excess energy is removed from the body by sweating, determine the volume of bodily fluid (assume it is water) lost per hour. (At 37.0 degree C the latent heat of vaporization of water is 2.41 Times 106 cm3

Explanation / Answer

so the energy we are interested in is the amount of energy each hour that goes into sweating

that is 91% of 355W

so that amount of energy per hour is 0.91x355J/s x 3600s=1.16x10^6J

this implies a mass of evaporated water of 1.16x10^6J/hr / 2.41x10^J/kg = 0.48 kg/hour

so, apart from using 91% vs. 9%, we are doing the same thing...now is the simplest part of the problem

the density of water is 1000kg/m^3, so this mass of water is the equivalent to a volme of

volume = mass/density = 0.48kg/1000kg/m^3 = 4.8x10^(-4)m^3 or 480 cm^3 or 0.48L

now, this assumes that sweat has the same density as fresh water which is probably not perfectly accurate but not a terrible approximation most likely

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