Question

2. An average human has about 67% water by mass. A 75.0 kg person suffering from hypothermia has a body water temperature of 33.5°C and needs to raise this to the normal body temperature of 37.0°C. a) How many grams of sucrose, C12H220(S), must be metabolized (use this as a simple combustion reaction with gaseous oxygen, producing liquid water and gaseous carbon dioxide)? g sucrose b) What volume of air (L) with a 1 atm total pressure and a partial pressure of oxygen of 0.2 atm would be needed for this? [Assume the air is 298 K.] b) Lair

Answer 1

mass of water = 75   $\times$$\frac{67}{100}$

= 50.25 Kg.

= 50.25 $\times$1000

= 50250 g

Heat of combustion of Sucrose = - 5640 KJ/mol.

heat released by combustion of 1 mole sucrose = 5640 KJ

now heat absorbed by water to increase the temperature from 33.5 to 37⁰ c.

q = mass of water $\times$ specific heat of water $\times$ ( temperature change)

= 50250 $\times$ 4.184 $\times$ (37 - 33.5)

= 735861 J

= 735.861 KJ

therefore moles of sucrose needed

= ( 735.861/5640)

= 0.130

mass of sucrose = moles $\times$ molar mass

= 0.130 (mol)  $\times$ 342 (g/mol)

= 44.5 g.

a)

balanced equation is

C₁₂H₂₂O₁₁ (s) + 12 O₂ (g)   $\to$   12 CO₂ (g) +11 H₂O (l)

mole ratio of Sucrose and oxygen = 1 : 12

then moles of oxygen needed = 12 $\times$ 0.130

= 1.56

now

Pressure of oxygen (P)= 0.2 atm

T = 298 K

number of moles (n ) = 1.56

R = 0.082 L-atm /mol.K

ideal gas equation is

PV =nRT

volume (V) = (nRT/P)

= (1.56 $\times$ 0.082 $\times$298 )/ 0.2

= 190.6 L

as total pressure of Air = 1 atm

and partial pressure of Oxygen = 0.2 atm

partial pressure of oxygen = Total pressure  $\times$ volume percentage  of oxygen

then percentage of oxygen = 20 %

therefore volume of air = (190.6$\times$100)/20 = 953 L air