4. An average man (mass = 70 kg, specific heat - same as water) produces about 1

Question

4. An average man (mass = 70 kg, specific heat - same as water) produces about 10^4 kJ of heat each day through metabolic activity.

i) If he were an isolated system, what would his temperature rise be in one day?

ii) He is, of course, really an open system--losing heat through evaporation of water. How much water must he evaporate per day to maintain his constant temperature of 37°C? You must first calculate Hvap at 37°.

iii) The heat of combustion of cane sugar is 3.95 kcal/g. How many grams of sugar will furnish energy for one day’s metabolism, assuming the transfer of heat from cane sugar bonds to metabolic heat to be perfectly efficient?

Explanation / Answer

m = 70 kg

Cp = 4.18

Q = 10^4 kJ/day

i) If he were an isolated system, what would his temperature rise be in one day?

Q = m*Cp*(Tf-Ti)

10^4kJ = 70 kg * 4.18 kJ/kgC * (Tf-37)

Tf = 71.2 °C

ii) He is, of course, really an open system--losing heat through evaporation of water. How much water must he evaporate per day to maintain his constant temperature of 37°C? You must first calculate Hvap at 37°.

Hvap at 37° = 2418 kJ/kg (from data bases)

then

Q = m*dHvap

10^4 kJ = m*2418 kJ/kg

m = 10^4 ) /2418) = 4.14 kg of water

iii) The heat of combustion of cane sugar is 3.95 kcal/g. How many grams of sugar will furnish energy for one day’s metabolism, assuming the transfer of heat from cane sugar bonds to metabolic heat to be perfectly efficient?

Hcomb = 3.95 kcal/g = 16526.8 J/g = 16.53 kJ/kg

Q = 10^4 kJ

Q = m*H

10^4 kJ = m * 16.53 kJ/kg

m = (10^4) kJ / (16.5 kJ/kg) = 606.1 kg

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