Modern Engineering Thermodynamics

710 CHAPTER 17: Thermodynamics of Biological Systems
IS IT A Calorie OR A calorie?
When the word calorie is capitalized, it indicates what nutritionalists call a large calorie, or a kilocalorie. That is, in nutrition
jargon, 1 Calorie = 1 kilocalorie, yet it takes 1000 calories to equal 1 kilocalorie. Only the capital C distinguishes the two.
So, when a nutrition table indicates that your caloric intake should be 2500 Calories per day, it really means 2500 kilocalories
per day. Now, since 1 kilocalorie = 4.186 kilojoules, then 2500 Calories/d = 2500 kcal/d = (2500 kcal/d)(4.186 kJ/
kcal) = 10,465 kJ/d = 10.456 MJ/d.
From Table 17.5, we find that the energy expenditure required to jog, play football, or fast dance is about 600 Calories per
hour, and Table 17.4 tells us that the energy content of milk chocolate is about 150 Calories per ounce. So, if you want to
exercise off the energy content of one 1.5-oz milk chocolate candy bar you would have to jog, play football, or fast dance
continuously for
This is a lot of hard exercise for one small candy bar.
ð1:5ozÞð150 Calories/ozÞ
600 Calories/h
= 0:375 hours
EXAMPLE 17.6
Suppose you want to exercise off the energy added to your body as a result of eating one pint of ice cream by lifting
weights. The external work done by the body equals the change in potential energy of the weights as they are lifted (there is
no significant energy recovery within the body, however, when the weights are lowered again). Suppose you are lifting 490.
N (110. lbm) a vertical distance of 1.00 m, and you can make one lift in 1.00 second. Approximately how many lifts are
required and how long will it take to work off the energy content of the ice cream?
Solution
Each lift requires that an amount of energy be put into the weights of
ðmgZ/g c
Þ weights
= ð490: NÞð1:00 mÞ/1
ð Þ = 490: N.m = 490: 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 O 2
and CO 2 exchanges), then we can write
_Q − _W =
dU + d
mV 2
+ d
mgZ
dt dt 2g c dt g c
Since the kinetic and potential energies of the human body do not change significantly during the exercise, we can set
and the energy rate balance becomes
d mV 2
dt 2g c
+ d dt
_Q − _W =
dU
dt
Now, the external work rate that must be done by the system is
mgZ
g c
body
body
= _U body
= 0
body
ð
_W = − mgZ/g cÞ weights
= 490: J = 490: J=s
Δt 1:00s
It has been shown experimentally that the energy conversion efficiency of animal muscular contraction defined by
Eq. (17.12) is about 25%, or
ðη T Þ muscle
=
_W
_U body
= 0:250