Question

The human body is about 2/3 water. Sucrose is metabolized by the body in the following reaction: C12H22O11 (s) + O2(g) --> H2O (l) + CO2 (g)

a. If a 63-kg person was out in the cold and got hypothermia, what mass of sucrose would need to be metabolized to raise the body temperature from 34.7 degrees Celsius to 37 degrees celsius? Assume products of metabolism are at 25 degrees Celsius.

b. If the oxygen gas has a partial pressure of 162 Torr, what volume (at 37 degrees Celsius) is required for the metabolism reaction?

Answer 1

It is a complicate question with missing some of the information.
In the first part you have to calculate the Energy that you need to change the temperature from 34.7 to 37C, with the formula
E=m x Ce x delta T

The mass I will assume that the problem want you to calculate the 2/3 of the total mass of the 63kg person that is water. And the Ce to use in your formula will be of the water.Water (liquid): cp = 4185.5 J/(kg·K) (15 °C, 101.325 kPa).

Then with the chemical equation of the combustion of the glucose you have to look in a table what is the delta H of the reaction (the delta H tables most of the time are a 25C ΔH°comb = -2801 kJ/mol of glucose) . With the amount of energy from the previous calculation and the delta H of the reaction you can calculate how many moles of glucose you will need. Knowing the moles easily you can calculate the grams (moles x Molecular mass of glucose).

For the second part of your question I would use the Ideal Gas General Law
PV=nRT knowing that you will need one mol of O2 per mol of glucose and the temperature has to be converted to Kelvin and the pressure to atmospheres, and R=0.082 atm L/mol K