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

[qspartan570]

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= $\frac{1}{2}$(70)(.30)²= 3.15

e$\frac{3.15}{1300}$=0.0024 X 100=0.024 %

According to the book the answer is 14%

How do I get to this answer?

 

[Andrew Mason]

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