Modern Engineering Thermodynamics

17.9 Thermodynamics of Aging and Death 719
absolute temperature T. A plot of ln k d vs. ln T yields a straight line from which the following equation can be
obtained:
k d = αT exp
s d
R − h d
RT
(17.30)
where α is a constant, T is the absolute temperature, and s d and h d are the specific molar activation entropy and
enthalpy of death. It has been shown from data on the death of unicellular organisms and the irreversible thermal
denaturization of proteins that s d and h d are related by
s d = h d
T c
+ β (17.31)
where T c = 330. K and is called the compensation temperature, and β is a constant equal to –276 kJ/(kgmole · K).
Equation (17.31) is often called a compensation law, because changes in s d are partially compensated for by
changes in h d , resulting in a relatively constant value of k d . The compensation is exact at T = T c as can be seen
by substituting Eq. (17.31) into Eq. (17.30). For the common housefly, the data reduction gives h d ≈ 800 MJ/
kgmole. Since k d is the death rate constant, the smaller it is, the smaller the death rate becomes and the longer
the life span becomes.
Combining Eqs. (17.30) and (17.31) gives
k d = αT
exp h d
R
T − T c
T × T c
+ β R
(17.32)
Now, because h d is such a large value and T < T c for most living systems, a small decrease in T can produce a
significant decrease in k d .UsingR = 8:3143 kJ/ðkgmole.KÞ, h d = 800,000 kJ/kgmole, β = –276kJ/ ðkgmole.KÞ,and
T c = 330. K, Eq. (17.32) becomes
n h
k d
α
= T exp 9:62 × 10 4 T − 330: io
− 33:2
330: × T
(17.33)
When T = 310. K (37.0°C), Eq. (17.33) gives k d /α = 8.03 × 10 –21 K. But when T is lowered by just 2 degrees to
308 K (35.0°C), then k d /α drops by almost a factor of 8 to 1.06 × 10 –21 K. If T is dropped all the way down to
293 K (20.0°C), then k d /α = 1.15 × 10 –28 K, a drop of a factor of 10 7 . Thus, the death rate constant is very sensitive
to the body temperature. It has been estimated by some researchers that, if the core temperature of humans
were lowered from its present value of 37°C down to 31°C, then the average age at death would increase from
about 75 to around 200 years.
EXAMPLE 17.10
The death rate constant for mice at 27.0°C is 0.0350 months –1 . Determine the coefficient α in Eq. (17.33) for mice.
Solution
At T = 27.0°C = 300. K, we find, from Eq. (17.33), that
k d /α = 300: exp 9:62 × 10 4
ð300: − 330: Þ/ ½ð330:
Þð300:
ÞŠ− 33:2 = 2:50 × 10 −25 K
then,
α = 0:035 months –1 /2:50 × 10 − 25 K = 1:4 × 10 23 months –1 .K −1
Exercises
28. Using the value for the coefficient α determined in Example 17.10, determine the death rate constant k d for mice at
30.0°C (303 K). Answer: k d = 0.846 months –1 .
29. Integrate Eq. (17.29) to find an expression for the ratio of the number of animals surviving at time t to the initial
number of animals present, N/N 0 . Then, using the information given in Example 17.10, evaluate this expression to
determine the percentage of mice that survive to 5.0 months at 27°C. Answer: N/N 0 × 100 = 84%.
30. Plot log(k d /α) vs. T for 30.0°C (303 K) ≤ T ≤ 40.0°C (313 K). Answer: See Figure 17.15.
(Continued )