2. a) Calculate the amount of water (in liters) that would have to be vaporized

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.

Explanation / Answer

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)

109*4.18* delT= 1250000,delT=0.000299

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