Overall 80 of the energy used by the body must be eliminated...

Overall, 80% of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation, evaporation of sweat (2,430 kJ/kg), evaporation from the lungs (38 kJ/h), conduction, and convection. A person working out in a gym has a metabolic rate of 2,500 kJ/h. His body temperature is 37°C, and the outside temperature 23°C. Assume the skin has an area of 2.0 m2 and emissivity of 0.97. (σ = 5.6696 10-8 W/m2 · K4) (a) At what rate is his excess thermal energy dissipated by radiation? (Enter your answer to at least one decimal place.) W (b) If he eliminates 0.44 kg of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (Enter your answer to at least one decimal place.) W (c) At what rate is energy eliminated by evaporation from the lungs? (Enter your answer to at least one decimal place.) W (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

Transcript

00:01 In the sum we want to determine the rate at which excess thermal energy dissipated by radiation second one, the rate at which excess thermal energy dissipated by evaporation of sweet a person eliminates 0 .41 kg of precipitation during that r.

00:18 Then third one, see here, the rate at which exists thermal energy eliminated by evaporation from the lungs.

00:26 And the last one is we want to find out the rate at which is the rate of which is, excess thermal energy emitted through conduction and the conduction.

00:35 So the given value of the sum is first one, metabolic rate of person working out in a gym so which is referred by the letter of q which is equal to 2 ,500 kj dividing hour, then temperature of the body that is t b is considered as the technical of the body which is equal to 37 degrees celsius which is equal to 310k okay then next temperature of outside which is considered by the letter of t o is the temperature of outside which is equal to 28 degrees celsius which is equal to 301 k next area of in of a body a which is equal to 2 meter square then mc which is of body which is e which is equal to 0 .97 then it's latent heat of evapation of sweet that is l sweet which is equal to 2 ,430 k j divided kilogram then next rate of evaporation from the lungs that is t lungs okay p of base lungs which is equal to 38 k -j dividing r.

02:00 Okay, so the next step is the explanation part that is step number two.

02:13 So first, a part is the expression of rate at which excess thermal energy dissipated by radiation is given by we have one type of formula we can use it now.

02:29 P radiation, which is hibati, equal to sigma a e multiple of t b power 4 minus t nr power 4 here sigma is the steep and b b b constant and a is the area e is the inextivity t b is the temperature of the body and t not is the temperature of outside next we want to substitute the value in the above expression that is p rickney of radiation which is equal to 5 .6696 multiple 10 to the power of minus 8 then multiple 2 multiple 0 .97 so multiple of 310 the whole power 4 minus 301 the whole power 4 okay so p radiation which is which is equal to 0 .0.

03:45 113 w that's the rate at which excess thermal energy dissipated by radiation is 0 .11 w okay then next step number three for this b partners the expression of rate at which excess thermal energy dissipated by evaporation of sweet is given by so we have one type of formula that formula is p -s -sweet which is equal to p -sweet which is equal to p -sweet which is equal to m, l -sweet dividing dill -t.

04:24 Okay, so here m is the mass of the perception and the i -sweet is the latin k, heat of evaporation of heat and dil t is the time.

04:35 So, such with the value in the above expression, we get the p -sweet value first.

04:41 Therefore, p -sweet which is equal to 276...

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