Heat engine/max height person can climb

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To estimate the maximum height a hiker can climb in one day using 4000 kcal, treat the hiker as a heat engine operating between an internal temperature of 37°C and an ambient temperature of 20°C. The efficiency of the heat engine can be calculated using the formula e = 1 - T(low)/T(high), which helps determine how much energy can be converted into useful work. The work required to climb is primarily the change in potential energy needed to overcome gravity. Understanding these principles allows for a rough calculation of the maximum height achievable. This approach provides a framework for solving the problem effectively.
Bradracer18
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Ok guys, I can't seem to get started on this one. A little starter would be nice if you can...and I'll try from there.


Assume that a hiker needs 4000 kcal of energy to supply a day's worth of metabolism. Estimate the maximum height the person can climb in one day, using only this amount of energy. As a rough prediction, treat the person as an isolated heat engine, operating between the internal temperature of 37 deg C and the ambient air temperature of 20 deg C.

Thanks,
Brad
 
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Do you know the formula for the efficiency of a heat engine?
 
e = 1 - Q(low)/Q(high)


I think that is it...but don't know how this helps me.
 
Replace those Q's by T's, and you have what you need. Then you know what fraction of the stored energy can be put into useful work. All you need then is to know how much work it takes to go up a certain height. They probably want you to just use the work it takes to overcome gravity, ie, the potential energy change (although realistically, this is only a small fraction of the total work done, otherwise it would be possible to walk on a flat plane for miles without burning a calorie).
 
OK...I'll try that and see what I come up with...thank you, you've been very helpful!

Brad
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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