Throughout our lectures we have learned that our body\'s primary defense against

Question

Throughout our lectures we have learned that our body's primary defense against high temperatures is through sweating (perspiration). Let's consider a 200 lb (91 kg) male farmer in the Central Valley of California on a July day when the air temperature is 110 degree F (43 degree C).On this day, the humidity is low enough that all the sweat evaporates from the farmer's skin, enabling him to maintain a body temperature of 100 degree F (38 degree C). The farmer sweats 1 liter (1000 g) per hour and 60% of the latent heat needed to evaporate this sweat is provided by farmer's body (the other 40% is provided by the air). If the farmer suddenly dehydrates and stops sweating, how long will it take for his body temperature to rise to a near-total 110 degree F (43 degree C) Assumptions you will need to use: 1. It takes 4200 Joules of energy to warm each kilogram (kg) of his body 1 degree C (1.8 degree F). 2. The latent heat of vaporization is 2.400,000 J/kg. This value means that it takes 2400,000 Joules of energy to convert one kilogram of sweat (a liquid) into a gas.

Explanation / Answer

Given 4200 joules per Kg is required for 1 C ( 1.8 F) rise in body temperature
Mass of man, m = 91 kg
so, energy required to raise his body temperature by (43 - 38) = 5 C = 91*4200*5 = 1,911,000 J

sweat = 1 kg per hour
latent heat required to evaporate 1 kg sweat = 2,400,000 J
heat provided by body per Kg sweat = 60 p.c. of latent heat = 1,440,000 J
heat taken by swat in t hours = 1,440,000t J
so, 1,440,000t = 1,911,000
t = 1.327 hours

so it would take 1.327 hours for the farmer's body temperature to reach fatal lim it if he stops swating

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