(21-23) thermodynamics problem. Finding efficiency for the human body

  • #1

Homework Statement



While running up stairs at a (vertical) rate of 0.30 m/s, a man of 70 kg generates waste heat at a rate of 1300 J/s. What efficiency for the human body can you deduce from this?



Homework Equations



efficiency of heat engine=e=[itex]\frac{W}{Q1}[/itex]=[itex]\frac{Q1-Q2}{Q2}[/itex]=1-[itex]\frac{Q2}{Q1}[/itex]



The Attempt at a Solution



Kinetic energy= [itex]\frac{1}{2}[/itex](70)(.30)2= 3.15

e[itex]\frac{3.15}{1300}[/itex]=0.0024 X 100=0.024 %

According to the book the answer is 14%

How do I get to this answer?
 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement



While running up stairs at a (vertical) rate of 0.30 m/s, a man of 70 kg generates waste heat at a rate of 1300 J/s. What efficiency for the human body can you deduce from this?
Your formula for efficiency is correct: η = W/Qh = Qh-Qc/Qh

Use W = Qh - Qc to find Qh from W and Qc.

Note: The kinetic energy of the man is irrelevant since it is not changing. The only work being done is in raising the mass of the runner through a vertical height. That is W. You are given Qc so find Qh. Then apply the formula for η.

AM
 

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