Modern Engineering Thermodynamics

17.6 Thermodynamics of Nutrition and Exercise 705
where BMR is in kJ/d and m is in kg. Then, the BMR per unit mass of an 80.0 kg human is
and the BMR per unit mass of an 8.00 gram mouse is
ðBMR/mÞ human = 293ð80:0 −0:25 Þ = 98:0
kJ
kg.d
ðBMR/mÞ mouse
= 293ð0:00800 −0:25 kJ
Þ = 980:
kg.d
This calculation shows that the basal metabolic rate per unit mass of a mouse is ten times that of a human. This large difference
is primarily due to the difference in surface area to volume ratio of these mammals. Since heat loss from the body is
primarily by convection heat transfer, which is proportional to surface area, and internal heat generation inside the body is
proportional to its volume, then as the ratio of surface area to volume increases the internal heat generation rate must also
increase if a mammal is to maintain its body temperature. To produce higher internal heat generation rates, small animals
must feed very often if they are not to starve.
It is a fact that the smaller any object becomes, the larger its surface area to volume ratio becomes. This is easiest to understand
with spherical objects. The surface area of a sphere is 4πR 2 whereas its volume is (4/3)πR 3 . Therefore, its surface area
to volume ratio is
Surface area
¼ 4πR2
Volume sphere 4
¼ 3 3 πR3 R
and this ratio decreases inversely with increasing R. Thus, there is a lower limit to the size of warm-blooded animals. The
shrew and the hummingbird are the smallest known animals of this kind.
The body temperature of insects and cold-blooded animals is approximately equal to the temperature of their surroundings.
Consequently, there is no thermodynamic lower limit to their size.
Exercises
7. Determine the basal metabolic rate (BMR) and the basal metabolic rate per unit mass (BMR/m) of a 1300. kg elephant.
Answers: BMR elephant = 63,400 kJ/d and (BMR/m) elephant = 48.8 kJ/kg · d.
8. If a 0.500 gram house fly had to maintain the body temperature of warm-blooded mammal using the same internal
heat generation rate mechanisms, what would be its basal metabolic rate (BMR) and its basal metabolic rate per unit
mass (BMR/m)? Answers: BMR fly = 0.980 kJ/d and (BMR/m) fly = 1960 kJ/kg · d.
9. Since whales are aquatic mammals, determine the basal metabolic rate (BMR) and basal metabolic rate per unit mass of
a 136,000 kg (150. ton) great blue whale. Answer: BMR whale = 2,080,000 kJ/d and (BMR/m) whale = 15.3 kJ/kg · d.
HOW DOES TEMPERATURE AFFECT METABOLISM?
Temperature governs metabolism through its effects on rates of biochemical reactions. Reaction kinetics vary with temperature
according to Boltzmann’s factor e − E/kT , where T is the absolute temperature, E is the activation energy, and k is Boltzmann’s
constant. The combined effects of body mass (M) and body temperature (T) on the basal metabolic rate can then
be written as BMR ∝ M 3/4 e − E/kT , where E is the average activation energy for the enzyme-catalyzed biochemical reactions of
metabolism.
17.6 THERMODYNAMICS OF NUTRITION AND EXERCISE
The molecular form of the food we eat can be broken down into the following three categories:
1. Carbohydrates. Carbohydrates always contain hydrogen and oxygen atoms in a 2 to 1 ratio, as in water; and
they can have very large macromolecules built up from the glucose (C 6 H 12 O 6 ) monomer, with molecular
masses as high as 2 × 10 6 (as in the case of plant starch and glycogen).
2. Proteins. Proteins are very large molecules containing carbon, hydrogen, oxygen, and often nitrogen. For
example, a single molecule of human hemoglobin (C 3032 H 4816 O 872 N 780 S 8 Fe 4 ) contains a total of 9512 atoms
and has a molecular mass of 66,552 kg/kgmole.
3. Fats (glycerol and fatty acids). Fatty acids are much smaller molecules, with typically 16 or 18 carbon
atoms per molecule plus attached hydrogen atoms and a carboxyl group (—COOH) at one end. An example
of a saturated (with hydrogen atoms) fatty acid is shown in Figure 17.8. An example of the same acid
unsaturated is shown in Figure 17.9.