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| Prime suspect (Posted on 2008-08-31) |
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You come into your professor's office to ask some
questions shortly before 9:00 a.m. on Friday.
You find him lying on
the floor of his office in a pool of chalk dust, dead.
You quickly
call the police and their investigators take several measurements over
the next hour, including:
1) the body temperature at 9:00 a.m. - 80 degrees.
2) the body temperature at 10:00 a.m. - 78 degrees
3) room temperature - 70 degrees (constant)
You realize that the police believe you to be a prime suspect,
so you need an alibi. You know that you were studying with friends until 3:00 a. m.,
but you aren't sure if that is enough information. You need to know
the time of death!
Assuming that the difference between body temperature and room
temperature changes at a rate proportional to that difference, and that the normal body temperature is 98.6 degrees, how
good is your alibi?
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Submitted by pcbouhid
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Solution:
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Nice work of Daniel and Dej Mar.
Based on the assumption in the last phrase, the equation, using T for temperature and
t for time is:
dT/dt = -k*(T - 70)
where k is the constant of proportionality.
So:
dT/(T - 70) = -k*dt
Integrating both sides:
ln(T - 70) = -k*t + constant
T - 70 = A*e^(-k*t) where A = e^(constant) = constant
At t = 9, T = 80 and at t = 10, T = 78.
So, 10 = A*e^(-9*k) and 8 = A*e^(-10*k)
Dividing these we have, 10/8 = e^(-9*k + 10*k) = e^k
This gives k = ln(5/4) = 0.223
To find A, use 10 = A*e^(-9 * .223) = 0.1342*A
So A = 10/.1342 = 74.5
The full equation is then: T = 70 + 74.5*e^(-.223*t)
Normal body temperature is 98.6, so now we require the value of t for when T = 98.6:
98.6 - 70 = 74.5*e^(-.223*t)
28.6/74.5 = e^(-.223*t)
.3839 = e^(-.223*t)
ln(.3839) = -.223*t
-.9574 = -.223t
t = .9574/.223
t = 4.293
This corresponds to 4:17 a.m., so you need to think of someone who can prove what you were doing at 4:17 a.m.
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