Asked by Bobby
An 80 kg hiker climbs a 1000 m high hill at a constant speed. If the hiker's body is 25% efficient, how many kg of fat does the hiker burn in climbing the hill? Given: heat of reaction of body fat is 37000 kilo joules per kilogram.
I know that change in potential energy is 800,000. I just don't know what to do next.
I know that change in potential energy is 800,000. I just don't know what to do next.
Answers
Answered by
Damon
m g h = increase of potential energy
80 * 9.81 * 10^3 = 785*10^3 Joules
785*10^3 = (1/4) total fuel energy burned
so
total fuel burned = 3.14 * 10^6 Joules
3.14 * 10^6 J / 37*10^6 J/kg = .085 kg
80 * 9.81 * 10^3 = 785*10^3 Joules
785*10^3 = (1/4) total fuel energy burned
so
total fuel burned = 3.14 * 10^6 Joules
3.14 * 10^6 J / 37*10^6 J/kg = .085 kg
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