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Modern Engineering Thermodynamics

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704 CHAPTER 17: <strong>Thermodynamics</strong> of Biological Systems<br />

WHAT IS KLEIBER’S LAW? Continued<br />

10 3<br />

10 −3<br />

10 −6<br />

10 0 10 −9<br />

Warm-blooded organisms<br />

Metabolic rate (kcal/h)<br />

Unicellular organisms<br />

Cold-blooded organisms<br />

10 −9<br />

10 −12<br />

FIGURE 17.7<br />

Kleiber’s law graph.<br />

10 −12 10 −6 10 −3 10 0<br />

Mass (grams)<br />

10 3 10 6 10 9<br />

WHAT DO A BANANA, AN ORANGE, AND A PERSON HAVE<br />

IN COMMON?<br />

According to a new study, all living organisms share roughly the same resting metabolic rate when body size and temperature<br />

are taken into account. The finding suggests that widely diverse species burn energy in predictable patterns. “The [corrected]<br />

basal metabolic rate of an apple or tree is remarkably similar to that of bacteria, which is remarkably similar to a<br />

fish or person,” says James Gillooly, at the University of New Mexico in Albuquerque.<br />

A comparison of the basal metabolic rates for a large number of warm-blooded animals from mice to elephants<br />

and birds produced the following empirical correlation, called Kleiber’s law:<br />

BMR = 293 m 3/4 (17.16)<br />

where BMR is the animal’s basal metabolic rate in kJ/d, and m is the animal’s body mass in kg. Thus, the basal<br />

metabolic rate per unit mass of the animal is<br />

BMR/m = 293ðm −1/4 Þ = 293<br />

m 1/4 (17.17)<br />

and it clearly increases with decreasing body mass.<br />

EXAMPLE 17.3<br />

Determine the basal metabolic rate (BMR) per unit mass of an 80.0 kg adult human and an 8.00 gram mouse. Since both<br />

are warm-blooded mammals, explain why there is a difference in these values.<br />

Solution<br />

The basal metabolic rate per unit mass of a warm-blooded animal is given by Eq. (17.17) as<br />

BMR/m = 293ðm −0:25 Þ

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