Question

Please answer with providing pictures. State any assumptions made and make sure to include all units with the answers and calculations! Thanks

You are studying for the first physics exam and forget to watch the time. The next morning you ignore your alarm clock until you realize you will not reach the bus unless you run as fast as you can to the bus station in 15min. Carrying your school books during the run results in a total energy use of 1200W. You start sweating, which is the way your body keeps you cool. Assume a typical efficiency how much would your body temperature rise during the 15min if you had no way of getting rid of your excess thermal energy? Assume that the only way that your body cools itself is evaporation of keep your body temperature the same, how much water do you have to evaporate during the 15 min of running? How much water do you need to drink not to get dehydrated? The heat of vaporization of sweat from skin is slightly a. 3 b. sweat. To higher than the value for water. Use the value given in your textbook. Possible enhancement (Warning, I have not calculated this, but one of you send an email reminding me to publish a HIP): What would be the outside temperature that you need to run in so you would radiate away the excess c. energy assuming that you would lose all extra heat through radiation, and your body temperature would be kept constant at 37°C Alternatively in case the numbers do not turn out reasonable: At what body temperature would you radiate away the excess energy assuming that you would lose all the extra heat through radiation. d.

Answer 1

heat generated from running in 15 mts

Q = 1200*15*60 J

we take specific heat of human body C = 4200 J/kg-K close to that of water

we take body weight M = 60 kg

C M T = Q

raise in temp T= Q/CM = 1200*15*60/4200*60 = 2.57 ^oC

b. heat of vaporisation of water = 2.25 E+6 J /kg

amount of water (sweat) to vaporise (kg) = 1200*15*60 / 2.25 E+6 = 0.48 kg

= 480 gms

to keep the body not de-hydrate we need to take 0.5 Ltr or more water.

c. Loss of heat from radiation

Q = e$\sigma$A( T⁴ -T_o⁴) per unit time

e - emisivity of the body we take as 0.5

$\sigma$ = 5.67E-8 W/m² - K⁴ - Stefan -Boltzman const.

A = 2 sq. total body surface area

T = 37 ⁰C = 309.89 K- typical human body temp.

T_o - external temp

time t = 15*60 = 900 s

The rate of emission =  0.5*5.67E-8*2(310⁴ -T_o⁴)

T_o = the rate of emission is 524 W approximately

The heat cannot be lost only by radiation at the normal body temp.

for the heat to loose by radiation we take external temp T_o = 300 K

1200 W = 0.5*5.67E-8*2(T⁴ -300⁴)

T = 413 K = 140.73 C - estimated body temp. to loose all heat by radiation.