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

## 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=$\frac{W}{Q1}$=$\frac{Q1-Q2}{Q2}$=1-$\frac{Q2}{Q1}$

## 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 Helper

## 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