Question

Calculate the enthalpy of breathing (at isobaric conditions) for a student that inhales air on a cool day (65F) and exhales air at body temperature (98.5F) , using the specific heat data shown in the table

The average volume of air in one breath is 230 cm³. The density of air is approximately 0.0012928 g/cm³. Estimate the amount of heat absorbed in one day by the air breathed by the student, assuming 10 breaths/min. Answer in units of KJ/day.

What happens if the same student inhales air on a hot summer day (100F) and exhales air at body temperature (98.5F)?

Specific heat at 1 atm; J/mol*k

Temperature, C	Specific heat
65	29.187
98.5	29.2369
110	29.2544

Answer 1

volume of air per breath = 230cm3

numbef of breaths/min =10, volume of total breaths =230*10 cm3/min

number of breaths/hr =60*230*10cm3/hr

number of breaths/day= 24*60*230*10cm3/day=3312000cm3/day

density = 0.0012928 g/cc. mass of air inhaled per day = 3312000* 0.001298 g/day=4282 g/day

moles of air = mass/molar mass = 4282/29 gmoles/day =148 gmoles/day

Enthalpy of air at 65deg.F assuming a reference temperature of 32deg.F =148*29.187*(65-32)=141636 J/day

enthalpy of air at 98.5 deg.F = 148*29.2369*(98.5-32)=287750 J/day

heat absorbed = 2788750-141636 J/day=141164 J/day=146.114 Kj/day

when the temperature is 100deg.F, the specific heat is considered as average of 98.5deg.F and 110 deg.F

specific heat = (29.2544+29.2369)/2=29.24 J/mole.deg.c

heat absorbed = 148*29.24*(100-98.5)=6491 J/day= 6.491 Kj/day