Question

2. a) Calculate the amount of water (in liters) that would have to be vaporized at 40°C (approximately body temperature) to expend the 2.5×10^6 calories of heat generated by a person in one day (commonly called sweating). The heat of vaporization of water at this temperature is 574 cal/g. We normally do not sweat that much. What is wrong with this calculation?

b)If you were in a reasonably well insulated camper, with a volume of about 10m^3, how much would the temperature increase over a 12 hour period?

Please explain in the greatest detail possible.

Answer 1

Heat of vaporization= 574 cal/g.

Calories of heat generated= 2.5*10⁶ calories

Mass of water required= 2.5*10⁶/574=4355.4gms

moles of water= 4355.4/18=242 moles

vapor pressure of water at 40 deg.c=55.3 mm Hg= 0.073 atm

at vaporization, vapor pressure= partial pressure = 0.073 atm

from PV= nRT

V= nRT/P= 242*0.08206*(40+273.15)/0.073=85188 L

when sweatting takes place the body temperatue drops down leading to reduction in temperature and further leading to reduction in extent of vaporization

b) volume of water=1000 m3= 1000*10³ L= 10⁶ L

density of water= 1000g/L   mass of water= 10⁶*1000 gms=10⁹ gms

heat gained in 12 hours perriods= 2.5*10⁶ calories/2= 1250000/12 hours

Mass* specific heat* (delT)= 1250000 (delT= temperature rise)

10^9*4.18* delT= 1250000,delT=0.000299