First Law Thermodynamics Problem

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Homework Help Overview

The discussion revolves around a thermodynamics problem related to the first law of thermodynamics, specifically focusing on an athlete's energy expenditure while pedaling on a gym bike. The problem involves calculating power, internal energy variation, and heat transfer based on the athlete's work and oxygen consumption.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conversion of energy units and the application of the first law of thermodynamics. There are questions about how to calculate internal energy variation and heat transfer based on the energy spent and work done. Some participants express uncertainty about the definitions and relationships between total energy, internal energy, and the work done.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and attempting to clarify the relationships between the various energy quantities involved. Some guidance has been offered regarding the application of the first law, but no consensus has been reached on the specific calculations.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to convert energy units and the implications of energy conservation principles. There is an acknowledgment of the assumptions regarding the athlete's physiological state during the exercise.

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Homework Statement


For an hour , an athlete with 50kg of weight produces a work of 75kcal pedaling on the gym bike. During this period of time he consumes 60L of oxygen.

Calculate:

a) The power produced by the athlete
b) The variation of the internal energy of the athlete knowing that for each liter of oxygen consumed, 5kcal of energy are spent.
c) The heat transferred to the envolving environment.

Homework Equations


First law of thermodynamics E2-E1=Q-W

The Attempt at a Solution



For the first one I converted the 75kcal that he produces in an hour to Joule/s giving me the power. Now, the other two I'm having trouble knowing what to do. If you could help me I would appreciate it.
Thank you,
 
Last edited:
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It's not cool to make duplicate posts of the same HW problem.
 
ruiwp13 said:
b) The variation of the internal energy of the athlete knowing that for each liter of oxygen consumed, 5kcal of energy are spent.
c) The heat transferred to the envolving environment.
b) Can you calculate how much energy was produced knowing how much oxygen was consumed?

c) I guess the athlete didn't develop a fever. So where did the energy go?
 
DrClaude said:
b) Can you calculate how much energy was produced knowing how much oxygen was consumed?

c) I guess the athlete didn't develop a fever. So where did the energy go?

in b) I know that 300kcal are spent (60L*5kcal) but that isn't the internal energy, I think,

in c) no, I guess he didn't :p and I know that he releases heat, but I'm not getting on the way to calculate the heat that he releases.
 
ruiwp13 said:
in b) I know that 300kcal are spent (60L*5kcal) but that isn't the internal energy, I think,
Indeed no, that is the total energy. You have to use the first law (conservation of energy) here.

ruiwp13 said:
in c) no, I guess he didn't :p and I know that he releases heat, but I'm not getting on the way to calculate the heat that he releases.
Solve b first, then c should be clear.
 
Last edited:
DrClaude said:
Indeed no, that is the total energy. You have to use the first law (conservation of energy) here.Solve b first, then c should be clear.

I know for c) I'll have to use the formula ΔE=Q-W, because I'm going to have the variation of internal energy and the work, but in b) When you say 300kcal is the total energy you mean it is ΔE=Ekinetic+Epotential+U?
 
ruiwp13 said:
I know for c) I'll have to use the formula ΔE=Q-W, because I'm going to have the variation of internal energy and the work, but in b) When you say 300kcal is the total energy you mean it is ΔE=Ekinetic+Epotential+U?
Use ΔE=Q-W for b).
 
DrClaude said:
Use ΔE=Q-W for b).

ΔE are the 300kcal and W are the 75kcal?
 
ruiwp13 said:
ΔE are the 300kcal and W are the 75kcal?
Correct. Actually, you only need the equation ΔE=Q-W, so sorry for the misdirection.
 
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DrClaude said:
Correct. Actually, you only need the equation ΔE=Q-W, so sorry for the misdirection.

No problem, yes, the heat is 225, in this case -225. And the variation is 300kcal, converting to Joules is 1255,2. Thank you! I'll try to find the variation now.
 

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