Newton's Law of Cooling

Date: 05/06/2003 at 18:45:54
From: Jenn
Subject: Newton's Law of Cooling

A famous math professor is found dead by his butler in his 50-room 
mansion. A student detective arrives on the scene and immediately 
measures the temperature of the body to be 87 deg F. After 3 hours the 
student measures the temperature of the body again and finds it has 
fallen to 84 deg F. Assuming the professor's original temperature was 
98.6 deg F, how long before the student arrived on the scene did the 
professor die?

(I-F)e^-kt+F
(98.6-?)e^-kt+F

Date: 05/06/2003 at 19:33:59
From: Doctor Luis
Subject: Re: Newton's Law of Cooling

Hi Jenn,

You almost have it with your two formulas

   (I-F)e^-kt+F
   (98.6-?)e^-kt+F

but you need to organize them better because they were taken at 
different times.

Newton's law of cooling states that the rate of cooling of an object 
is proportional to the temperature difference between the object and 
the environment. So, suppose the temperature of the professor after 
being killed is P(t), where t=0 is the time he was killed.

   dP/dt = k(R-P(t))

In this formula, k is the rate of cooling (in deg F/hour), R is the 
room temperature, which I'll assume to be 68 degrees F, although it 
typically varies from 68F to 77F.

So, solving for P in the equation above, we find

   dP/(P(t)-R) = -k dt

   ln(P(t)-R) = -kt + ln C

   P(t) = R + C e^(-kt)

So far, we don't know the constants C or k. But we do know three 
things:

   P(0 hours)   = 98.6 degrees F
   P(T hours)   = 87 degrees F
   P(T+3 hours) = 84 degrees F

where T is the time the professor was found by the student. This is 
also an unknown.

This gives us three equations that we can use to find our three 
unknowns. They are,

   98.6 = R + C
   87   = R + C e^(kT)
   84   = R + C e^(k(T+3)) = R + C e^(kT) e^(3k)

Recall that R = 68.

The first equation gives us the value of C. We can use the second 
equation, plus our new knowledge of C, to find the value of kT. And 
finally, we can use our knowledge of kT and C in the third equation to 
find the value of k. Once you know k, you can use the value of kT to 
find T.

The professor died about 8 hours 18 minutes before he was found 
(T=8.319). What is the value of C and k?

I guess I should add that if R = 77 degrees F, then solving for T 
again, you'd find that T = 6.477 hours (about 6 hours 28 minutes).
So there's an uncertainty of (8.319-6.477)hours = .921 hours = 55 mins
in the time of the professor's death. It would depend on the room
temperature over the time he's been dead. Unless you knew the
true time dependence R(t) of the room temperature and solving the
differential equation again, you're probably better off making
a high and a low estimate of R.

Date: 05/06/2003 at 19:51:00
From: Doctor Anthony
Subject: Re: Newton's Law of Cooling

If the ambient temperature of the room is @, then Newton's Law of 
Cooling gives the equation

     dT/dt = -k.(T - @)    where T = temperature at time t.

     dT
   ----- = -k.dt    and integrating
    T-@

    ln(T-@) = -kt + C

       T-@ = e^(-kt+C)

       T-@ = A.e(-kt)

at t=0  T = 87

      87-@ = A

      T-@ = (87-@).e^(-kt)

at t=3  T = 84

      84-@ = (87-@).[e^(-k)]^3      

      (84-@)
      ------ = [e^(-k)]^3
       87-@

so e^(-k) = [(84-@)/(87-@)]^(1/3)

When t = 98.6 we have

    98.6-@ = (87-@).e^(-k)^t

    98.6-@ = (87-@).[(84-@)/(87-@)]^(t/3)

We can solve this if we know the ambient temperature @.  If we assume 
the standard ambient temperature of 60 degrees F then we have

     38.6 = 27(24/27)^(t/3)

   1.42963 = (8/9)^(t/3)

Taking logs we get

    (t/3)ln(8/9) = ln(1.42963)

                   3.ln(1.42963)
              t =  -------------
                      ln(8/9)

                =  -9.1035

                =  9 hrs  6 mins  13 secs

So the professor died 9 hours 6 minutes before he was discovered by 
the student.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/