Hiker climbs top of mountain; energy conversion problem

[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?

 

[mgb_phys]

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

 

[Borek]

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

 

[pentazoid]

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?

 

[Borek]

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.

 

[pentazoid]

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)?

 

[Borek]

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

 

[pentazoid]

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?

 

[Borek]

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

 

[pentazoid]

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

 

[AROD]

my attempt

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

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

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?

 

[Borek]

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

 

[AROD]

gotcha, my bad.

physics rocks.