Expert: Paul Klarreich - 1/25/2007

Question
I asked this before (not to you) and I was not 100% with understanding the answer (understanding how to manipulate the equations.

On a hot day a thermometer is takend outside from an air conditioned room where the temperature is 21C. After one minute it reads 27 and after two it reads 30.  What is the temperature outside.

I know this is an exponential rate problem. It is just how to get from dT/dt=k(T-C) where T is the final temperature, C is the temperature of the surroundings and k is a constant to the answer.

I think once I understand exactly how they get from dT/dt=k(T-c) to T=uoe^kt (maybe there is a property of logs I am not seeing when they go from one for to another) and then where all these numbers go I will have a better grasp.

Hope that made sense...

Answer
Questioner:   Aaron
Category:  Calculus
 
Subject:  Newton's law of cooling...
Question:  I asked this before (not to you) and I was not 100% with understanding the answer (understanding how to manipulate the equations.

On a hot day a thermometer is takend outside from an air conditioned room where the temperature is 21C. After one minute it reads 27 and after two it reads 30.  What is the temperature outside.

I know this is an exponential rate problem. It is just how to get from dT/dt=k(T-C) where T is the final temperature, C is the temperature of the surroundings and k is a constant to the answer.

I think once I understand exactly how they get from dT/dt=k(T-c) to T=uoe^kt (maybe there is a property of logs I am not seeing when they go from one for to another) and then where all these numbers go I will have a better grasp.

Hope that made sense...
............................................
Hi, Aaron,

The basic scheme is your standard 'growth and decay' setup.  And you are right that:

dT/dt = k(T - C), where, I assume,

T is the reading on the thermometer, and
C is the outside temperature, which we don't know.

What you have there is a Differential Equation -- an equation involving one or more derivatives of an unknown function.  In this case, the unknown function is T(t) and we solve the equation by Separation of Variables.  

The variables are  T and t.  (Not T and C, even though you are supposed to find C.  For the purposes of this equation, C is a constant.)

After separating, the equation looks like this:
 dT
----- = k dt
T - C

and we integrate each side.

ln(T - C) = kt + A   (A constant of integration.)

Now exponentiate to get:

T - C =  e^(kt + A)

and rewrite based on exponent properties:

T - C =  e^kt e^A
T - C =  a e^kt   <<  a = e^A, which is a constant.
T = C +  a e^kt   

Now put in some of the 'observational' data.  Since you have three unknown constants, a, k, C, you need three observations.  You have:

At  t = 0,  T = 21.  
At  t = 1,  T = 27.
At  t = 2,  T = 30.

So you have three equations:

21 = C + a e^k(0)
21 = C + a          [I]


27 = C +  a e^k(1)   
27 = C +  a e^k     [II]


30 = C +  a e^k(2)    
30 = C +  a (e^k)^2    [III]


Here they are again:

21 = C + a          [I]
27 = C + a e^k      [II]
30 = C + a (e^k)^2  [III]

Now do some standard simultaneous equation stuff to eliminate:
II - I  gives:

6 = a e^k - a
a e^k = 6 + a
       6 + a
a e^k = ------
         a

Next, III - I  gives:

9 = a (e^k)^2 - a

     6 + a
9 = a(-----)^2 - a
       a
     36 + 12a + a^2
9 = a(--------------) - a
           a^2

    36 + 12a + a^2
9 = (--------------) - a
           a

9a = 36 + 12a + a^2 - a^2
9a = 36 + 12a

a = - 12

Finally, in equation I:

21 = C - 12           [I]

C = 33  degrees.