Question

Calculate the mass of glucose metabolized by a 78.8 kg person in climbing a mountain with an elevation gain of 1650 m . Assume that the work performed in the climb is four times that required to simply lift 78.8 kg by 1650 m .

Answer #1

Calculate the mass of glucose metabolized by a 82.1 kg person in
climbing a mountain with an elevation gain of 1910 m . Assume that
the work performed in the climb is four times that required to
simply lift 82.1 kg by 1910 m .

The metabolism of glucose, C6H12O6, yields carbon dioxide,
CO2(g), and water, H2O(l), as products. Energy released in this
metabolic process is converted to useful work, w, with about 63.0 %
efficiency. Use the data below to answer questions about the
metabolism of glucose. Calculate the mass of glucose metabolized by
a 75.5 kg person in climbing a mountain with an elevation gain of
1870 m . Assume that the work performed in the climb is four times
that required to...

The metabolism of glucose, C6H12O6, yields carbon dioxide,
CO2(g), and water, H2O(l), as products. Energy released in this
metabolic process is converted to useful work, w, with about 68.0 %
efficiency. Use the data below to answer questions about the
metabolism of glucose.
Substance ΔH∘f (kJ/mol)
CO2(g) −393.5
C6H12O6(s) −1273.3
H2O(l) −285.8
O2(g) 0
Calculate the mass of glucose metabolized by a 72.2 kg person in
climbing a mountain with an elevation gain of 1210 m . Assume that
the...

The metabolism of glucose, C6H12O6, yields carbon dioxide,
CO2(g), and water, H2O(l), as products. Energy released in this
metabolic process is converted to useful work, w, with
about 67.0 % efficiency. Use the data below to answer
questions about the metabolism of glucose.
Substance
ΔH∘f
(kJ/mol)
CO2(g)
−393.5
C6H12O6(s)
−1273.3
H2O(l)
−285.8
O2(g)
0
Part A
Calculate the mass of glucose metabolized by a 84.3 kg person in
climbing a mountain with an elevation gain of 1490 m . Assume that...

The metabolism of glucose, C6H12O6, yields carbon dioxide,
CO2(g), and water, H2O(l), as products. Energy released in this
metabolic process is converted to useful work, w, with
about 71.0 % efficiency. Use the data below to answer
questions about the metabolism of glucose.
Substance
ΔH∘f
(kJ/mol)
CO2(g)
−393.5
C6H12O6(s)
−1273.3
H2O(l)
−285.8
O2(g)
0
Part A
Calculate the mass of glucose metabolized by a 75.9 kg person in
climbing a mountain with an elevation gain of 1610 m . Assume that...

± Metabolism and Work
The metabolism of glucose, C6H12O6, yields carbon dioxide,
CO2(g), and water, H2O(l), as products. Energy released in this
metabolic process is converted to useful work, w, with
about 74.0 % efficiency. Use the data below to answer
questions about the metabolism of glucose.
Substance
ΔH∘f
(kJ/mol)
CO2(g)
−393.5
C6H12O6(s)
−1273.3
H2O(l)
−285.8
O2(g)
0
Part A
Calculate the mass of glucose metabolized by a 68.4 kg person in
climbing a mountain with an elevation gain of 1990...

The metabolism of glucose is about 66% efficient, meaning that
about 66% of the heat released during the breakdown of glucose is
converted to useful work. A 70.0 kg hiker climbs up a rocky outcrop
from an initial elevation of 1010 ft to a final elevation of 1634
ft. Calculate the mass of glucose (grams) metabolized during the
hike. (Assume the actual work performed by the hiker is five times
that required to lift 70.0 kg through an elevation from...

Can I have all the answer please....
1. A 75 kg man climbs a mountain 1000m high in 3hours and uses
9.8kcal/min. Calculate the energy consumed in Joules/sec?
2.
Which of the following are correct units for power? (Note: there
may be more than one answer here)
A.
Joules
B.
Joules/sec
C.
Newtom meter
D.
Newton/meter2
E.
Newton/(meters *second)
F.
Newton Meter/Second
G.
Watts
H.
3.
Consider the man from question number 1. What is his power
consumption in Watts?...

A person of mass 75 kg is standing on the surface of the
Earth.
a. Calculate the total energy of this individual, assuming they
are at rest relative to the surface. (Hint: Use the expression for
gravitational potential energy (*NOT Ug=mgh*)). Although
the person is standing at rest relative to the surface of the
Earth, they are rotating along with the Earth and thus store some
kinetic energy.) Ignore the rotation of the Earth about the Sun,
the motion of...

A mountain climber of mass m=75.0 kg slips during a climb and
begins to fall freely. She forgot her rope and had to use an ideal
spring. The spring is hanging vertically with an unstretched
equilibrium position at the origin, x=0. On her descent, she grabs
the end of the spring which has a spring constant of k= 981 N/m.
Ignore friction, air resistance, and any other dissipative
effects.
a) The climbers velocity when she grabbed the spring was 20.0...

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