Question 17.6: Suppose you want to exercise off the energy added to your bo...

Each lift requires that an amount of energy be put into the weights of

(mgZ/\text{g}_c)_\text{weights} = (490.  \text{N}) (1.00  \text{m})/ (1) = 490.  \text{N.m} = 490.  \text{J}

If we take the human body as the thermodynamic system and apply the energy rate balance and ignore all mass flow energy movements into or out of the system during the exercise period (thus, we are ignoring perspiration energy losses and all \text{O}_2 and \text{CO}_2 exchanges), then we can write

\dot{Q} − \dot{W} = [\frac{dU}{dt} + \frac{d}{dt} (\frac{mV^2}{2\text{g}_c}) + \frac{d}{dt} (\frac{m\text{g}Z}{\text{g}_c})]_\text{body}

Since the kinetic and potential energies of the human body do not change significantly during the exercise, we can set

[\frac{d}{dt} (\frac{mV^2}{2\text{g}_c}) + \frac{d}{dt} (\frac{mgZ}{\text{g}_c})]_\text{body} = 0

and the energy rate balance becomes

\dot{Q} − \dot{W} = (\frac{dU}{dt})_\text{body} = \dot{U}_\text{body}

Now, the external work rate that must be done by the system is

\dot{W} = – \frac{(m\text{g}Z/\text{g}_c)_\text{weights}}{Δt} = \frac{490.  \text{J}}{1.00 \text{s}} = 490.  \text{J/s}

It has been shown experimentally that the energy conversion efficiency of animal muscular contraction defined by Eq. (17.12) is about 25%, or

\text{Energy conversion efficiency} = η_E = \frac{\text{Desired energy result}}{\text{Required energy input}}                  (17.12)

(η_T)_\text{muscle} = \frac{\dot{W}}{\dot{U}_\text{body}} = 0.250

Then, the rate of total internal energy expenditure within the body is

\dot{U}_\text{body} = \frac{- \dot{W}}{(η_T)_\text{muscle}} = \frac{−490  \text{J/s}}{0.250} = −1960  \text{J/s}

Therefore, \dot{Q} = \dot{U} + \dot{W} = −1960 + 490 = −1470  \text{J/s}. Consequently, the time, τ, required to produce a change in the total internal energy of the system that equals the energy content of one pint of ice cream (see Table 17.4) is

τ = (\frac{ΔU}{\dot{U}})_\text{body} = \frac{- (1  \text{pint}) (2.51  \text{MJ/pint})}{−1.96 × 10^{−3} \text{MJ/s}} = 1280 \text{s} = 21.3  \text{min}

Hence, the 490. N weight in this example must be lifted continuously at a rate of one lift per second until a total of 1280 lifts have been made. This is clearly a great deal of physical labor just to overcome the enjoyment of a pint of ice cream.
Note that only 25% of the energy in the ice cream gets converted into external work while 75% of its energy is utilized elsewhere within the body to keep the circulatory, respiratory, and other subsystems operating and is ultimately converted into heat inside the body due to the internal irreversibilities of these processes.