# Hiker climbs top of mountain; energy conversion problem

• pentazoid
In summary, a 60 kg hiker will need to eat 8.43 bowls of corn flakes before setting out on a 1500 m ascent to Mt. Ogden, assuming she is 25% efficient at converting chemical energy and all mechanical work is used to climb vertically. As she climbs, 3/4 of the energy from the corn flakes will be converted to thermal energy, resulting in a 10.5 degree increase in body temperature if there is no dissipation. In reality, this energy is mostly used to evaporate water from her skin, leading the hiker to drink 1.09 liters of water to replenish lost fluids.
pentazoid

## Homework Statement

A 60 kg hiker wishes to climb to the summit of Mt. Ogden , an ascent of 1500 m.

a) Assuming that she is 25 % efficient at converting chemical energy from food into mechanical work, and that essentially all the mechanical work is used to climb vertically , roughly how many bowls of corn flakes(standard serving 1 ounce, 100 kilocalories) should the hiker eat before setting out?

b) As the hiker climbs the mountain , 3-quarters of the energy from the corn flakes is converted to thermal energy. If there were no way to dissipate this energy , how many degrees would her body temeperature increase?

c) In fact, the extra energy does not warm the hiker's body significantly; instead, it goes(mostly) into evaporating water from her skin. How many liters of water should she drink during the hike to replace the lost fluids?(At 298 K, a reasonable temperature toassume , the latent heat of vaporization of water is 580 cal/g, 8 % more than at 373 K)

## Homework Equations

mgh=U
Q=delta(T)*C
PV=RT
delta(H)=delta(U)+Pdelta(V)

## The Attempt at a Solution

a) mgh=(60)(9.8)(1500) =882000 joules=882 kJ

If she works at 25 % efficiency, I should considered only .75*mgh=662 kJ

4.184 kJ=1 kilocalorie => 662 kJ=158 kilocalories==> 1 .58 ounces or 1.58 bowls of corn flakes

b) delta(H)=Q+W_other. Does no dissipation mean Q=0? If so then delta(T) = 0

c) Q=Lm=(580 cal/g)(18 g)= 10440 g; PV=RT; T=298 K, R=8.31 J/K, P=1.01e5 , why would they give me the latent heat of vaporization when I can just T, R and P to find the volume?

pentazoid said:
a) mgh=(60)(9.8)(1500) =882000 joules=882 kJ
Looks good

If she works at 25 % efficiency, I should considered only .75*mgh=662 kJ
Nope, if the hiker is only 25% efficent you need 100/25 = 4x as much energy.

c) Q=Lm=(580 cal/g)(18 g)= 10440 g; PV=RT; T=298 K, R=8.31 J/K, P=1.01e5 , why would they give me the latent heat of vaporization when I can just T, R and P to find the volume?
You are evaporating liquid water into vapour

b - no dissipation measn no heat is lost to the surroundings, all 75% heats her body

c - you need latent heat to calculate amount of water that have to be evaporated to dissipate the heat

Borek said:
b - no dissipation measn no heat is lost to the surroundings, all 75% heats her body
c - you need latent heat to calculate amount of water that have to be evaporated to dissipate the heat

so Q=0 and therefore delta(T)=0 since Q=delta(T)*C

What quantity represents the amount of water? Q? I don't need the relation PV=RT, to find V?

pentazoid said:
so Q=0 and therefore delta(T)=0 since Q=delta(T)*C

No, Q is not 0.

What quantity represents the amount of water? Q? I don't need the relation PV=RT, to find V?

I think you are on the right track.

Borek said:
No, Q is not 0.

Wait, I thought no dissipation means no heat? should I used the fact that delta(U)=f/2*N*k*delta(T)?

No dissipation means no heat exchange, that's not the same as no heat produced.

Borek said:
No dissipation means no heat exchange, that's not the same as no heat produced.

delta(U)=f/2*N*k*Delta(T),f=5 since hiker is evaporating H2O

delta(U)=.75*mgh

therefore I can now find delta(T), right?

You are overcomplicating. You know hikers mass, assume specific heat of water.

Borek said:
You are overcomplicating. You know hikers mass, assume specific heat of water.

Ahh, of course, delta(U)-Q and I can find delta(T) from this relation since mechaniccal energy goes into 75 % thermal energy

my attempt

pentazoid said:

## Homework Statement

A 60 kg hiker wishes to climb to the summit of Mt. Ogden , an ascent of 1500 m.

a) Assuming that she is 25 % efficient at converting chemical energy from food into mechanical work, and that essentially all the mechanical work is used to climb vertically , roughly how many bowls of corn flakes(standard serving 1 ounce, 100 kilocalories) should the hiker eat before setting out?

a) mgh=(60)(9.8)(1500) =882000 joules=882 kJ

If she works at 25 % efficiency, she must do 4 times the work, or 4*mgh= 3528 kJ

4.184 kJ=1 kilocalorie => 3528 kJ= 843 kilocalories==> 8.43 ounces or 8.43 bowls of corn flakes

pentazoid said:
b) As the hiker climbs the mountain , 3-quarters of the energy from the corn flakes is converted to thermal energy. If there were no way to dissipate this energy , how many degrees would her body temeperature increase?

3/4 energy from the corn flakes is 3*882 = 2646 kJ

It takes 4.186 J to heat 1 kg water 1 degree.

(2646 kJ/60 kg)*1K*kg/4.186kJ = 10.5 degree increase

pentazoid said:
c) In fact, the extra energy does not warm the hiker's body significantly; instead, it goes(mostly) into evaporating water from her skin. How many liters of water should she drink during the hike to replace the lost fluids?(At 298 K, a reasonable temperature to assume, the latent heat of vaporization of water is 580 cal/g, 8 % more than at 373 K)

If 580 cal of energy will evaporate 1 g of water, the thermal energy she gives off 2464 kJ or 632 kcal will evaporate 632/.58 g = 1090 g = 1.09 L of water that she should drink

Or?

AROD, what you just did was you have done someone else's homerwork. Please don't.

Last edited:

physics rocks.

## What is the energy conversion problem in a hiker climbing to the top of a mountain?

The energy conversion problem in a hiker climbing to the top of a mountain refers to the transformation of energy from one form to another as the hiker moves up the mountain. This includes converting chemical energy from food into mechanical energy for movement, potential energy as the hiker gains height, and kinetic energy as the hiker moves upward.

## How does the hiker's body convert energy during the climb?

The hiker's body converts energy through various physiological processes. The digestive system breaks down food into glucose, which is then converted into ATP (adenosine triphosphate), the body's main source of energy. The muscles then use this ATP to perform work, such as movement, which requires the conversion of chemical energy into mechanical energy.

## What are the main sources of energy for a hiker?

The main sources of energy for a hiker are food and oxygen. Food provides the necessary nutrients and calories for the body to function and perform physical activities. Oxygen is also essential for energy production as it is required for the breakdown of glucose to produce ATP in the body's cells.

## Why does the hiker feel tired during the climb?

The hiker feels tired during the climb because the body's energy stores are being used up to power the physical activity. As the hiker continues to climb, the body must work harder to maintain the same level of movement, resulting in fatigue. Additionally, the body may also experience a build-up of lactic acid, which can contribute to feelings of fatigue.

## What is the role of energy conservation in the hiker's climb?

Energy conservation is crucial for the hiker's climb as it allows the body to efficiently use and manage its energy resources. This includes pacing oneself, taking breaks to rest and refuel, and maintaining proper hydration and nutrition. By conserving energy, the hiker can avoid exhaustion and successfully reach the top of the mountain.

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