Question

d) AE = . 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, C12H22O11 (s), must be metabolized (use this as a simple combustion reaction with gaseous oxygen, producing liquid water and gaseous carbon dioxide)? 42 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.] 4 g sucrose b) L air Brands x 290001, Dayton, Ohio 45429 mead.com 11233 02016 ACCO Brands ghts Reserved

Answer 1

Following is the - complete Answer: -&- Explanation: for the given: Question: in....typed & image format:

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$\Rightarrow$Part -(a):

$\Rightarrow$Answer:

Grams of sucrose ( C₁₂H₂₂O₁₁ ): required to be metabolized = 44.62 g ( grams )

$\Rightarrow$Explanation:

Following is the complete: Explanation: for the above: Answer: ..in...typed & image format...

• Given:

1. In a average human: percentage of water present = 67.0 %  ( i.e. percentage by mass )
2. Mass of the body of an average human = 75.0 kg = 75000.0 g ( grams )
3. Specific Heat Capacity Of Water:  C_w = 4.18 kJ/ ( kg. ^oC ) ...approx.
4. Initial Temperature of Human Body: T_i = 33.5 ^oC
5. Final Temperature of the Human Body:  T_f = 37.0 ^oC..

• ​​​​​​​Step - 1:

​​​​​​​Following is the parts of the Explanation: in.....image format...

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$\Rightarrow$Part -(b):

$\Rightarrow$Answer:

Volume of air:required = 166.4 L ( Liters )

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above: Answer...in...image format...

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