Respuesta :
Answer:
The mass of glucose metabolized by a 78.8 kg person in climbing a mountain with an elevation gain of 1570 m, is 311.5 grams
Explanation:
Step 1: Data given
Mass of the person = 78.8 kg
elevation = 1570 meters
Molar mass of glucose = 180.16 g/mol
Standard enthalpy of combustion of glucose = -2805 kJ/ mol
Step 2:
W = m*g*h
with m = mass of the person in kg = 78.8 kg
with g = acceleration due to gravity = 9.81 m/s²
with h = height in meters = 1570 meters
W = 78.8 * 9.81 * 1570 = 1213653.96 J = 1.213 *10^6 J
Step 3: Calculate actual work
Actual work = 4W = 4.85 *10^6 J = 4.85 *10³ kJ
Step 4: Calculate moles glucose
Moles glucose = work / standard enthalpy
Moles glucose = 4.85 *10³ kJ / 2805 kJ/mol = 1.729 moles
Step 5: Calculate mass of glucose
Mass glucose = moles glucose * Molar mass glucose
Mass glucose = 1.729 moles * 180.16 g/mol
Mass glucose = 311.5 grams
The mass of glucose metabolized by a 78.8 kg person in climbing a mountain with an elevation gain of 1570 m, is 311.5 grams
The mass of glucose metabolized by a 78.8 kg person in climbing a mountain with an elevation gain of 1570 m, is 311.5 grams
Calculation of mass of glucose:
Since
Mass of the person = 78.8 kg
elevation = 1570 meters
Molar mass of glucose = 180.16 g/mol
Standard enthalpy of combustion of glucose = -2805 kJ/ mol
Now
We applied the below formula
[tex]W = m\times g\times h[/tex]
Here
m = mass of the person in kg = 78.8 kg
g = acceleration due to gravity = 9.81 m/s²
h = height in meters = 1570 meters
So,
[tex]W = 78.8 \times 9.81 \times 1570[/tex]
= 1213653.96 J
[tex]= 1.213 \times 10^6 J[/tex]
Now actual work is
= 4W
[tex]= 4.85 \times 10^6 J \\\\= 4.85 \times 10^3 kJ[/tex]
Now moles glucose is
Moles glucose is
[tex]= work \div standard\ enthalpy\\\\ = 4.85 \times 10^3 kJ \div 2805 kJ/mol[/tex]
= 1.729 moles
Now the mass of glucose
[tex]Mass glucose = moles\ glucose \times Molar\ mass\ glucose\\\\= 1.729 moles \times 180.16 g/mol[/tex]
= 311.5 grams
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