Newton's Law of cooling? HELP?????

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klooless

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Jun 10, 2009
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This problem is the law just backwards. There are also two unknown constants and i therefor have no idea how to manipulate the equation when two are missing...

Question:
On a hot day a thermometer is taken outside from an air-conditioned room where the temperature is 21 degrees C. After one minutes it reads 27 degrees C, and after two minutes it reads 30 degrees C. Use Newton's law of cooling to find the outdoor temperature.

My work so far:
dy/dx = K(y-?) ?=outdoor temperature y=thermometer reading

21=(21-?) e^K(0)
27=(27-?) e^K(1)
30=(30-?) e^K(3)


What should I do??? I really need guidance on this problem!!!!!!!!
 
In this case, we have three variables to solve for instead of two. That's the only difference. That's enough, huh :D

Start with the Newton formula for cooling:

\(\displaystyle \frac{dT}{dt}=k(T-T_{m})\), where \(\displaystyle T_{m}\) is the outside temperature.

Separate:

\(\displaystyle \frac{dT}{T-T_{m}}=kdt\)

Integrate

\(\displaystyle \int\frac{dT}{T-T_{m}}=\int kdt\)

\(\displaystyle ln(T-T_{m})=kt+C\)

\(\displaystyle T-T_{m}=C_{1}e^{kt}\)

\(\displaystyle T=T_{m}+C_{1}e^{kt}\)

Now, here is where we use our conditions.

Start with the initial condition T(0)=21.

\(\displaystyle 21=T_{m}+C_{1}e^{k(0)}\Rightarrow 21=T_{m}+C_{1}\)

\(\displaystyle T_{m}=21-C_{1}\)

Sub back into the equation:

\(\displaystyle T=(21-C_{1})+C_{1}e^{kt}\)

Now,we're down to two variables:

Next, use the other initial condition, T(1)=27.

\(\displaystyle 27=(21-C_{1})+C_{1}e^{k(1)}\)

\(\displaystyle C_{1}=\frac{6}{e^{k}-1}\)

Sub back in and use the third I.C.

\(\displaystyle 30=(21-\frac{6}{e^{k}-1})+\frac{6}{e^{k}-1}\cdot e^{k(2)}\)

Now, k is the only variable left, so we solve for k and we find \(\displaystyle \boxed{k= -ln(2)}\)

Back sub to find \(\displaystyle C_{1}=\frac{6}{e^{-ln(2)}-1}=-12\)

\(\displaystyle \boxed{T_{m}=21-(-12)=33}\)

There are all the variables and we have:

\(\displaystyle \boxed{T=33-12e^{-ln(2)t}}\)

This ought to help you in future Newton problems. If it is the easier two variable version, do the same only omit a step.
 
\(\displaystyle Good \ show \ galactus, \ as \ I \ followed \ your \ analysis \ and \ I \ too \ got \ T(t) \ = \ 33-\frac{12}{2^{t}}.\)

\(\displaystyle Hence, \ the \ outside \ temperature \ must \ be \ 33C \ which \ boils \ down \ to \ 91.4F \ a \ hot \ one.\)
 
BigGlenntheHeavy said:
\(\displaystyle Hence, \ the \ outside \ temperature \ must \ be \ 33C \ which \ boils \ down \ to \ 91.4F \ a \ hot \ one.\)
Click to expand...

Not in Arizona - or anywhere in India..... :lol: :lol: :lol:

By the way, that thermometer should be thrown away - after 2 minutes it is still ~10% off !!! Even my Dell_PC boots up before that !!!
 
Subhotosh Khan, I was in Arizona in the summer time and if you didn't have air conditioning, you woke up to a soggy mattress. The reason there are so many Indians in Arizona is when the white man went through there in the 1800's, he said "let them have it (the Indians), I'm out of here".

In fact, to this day I remember living in an air condition motel and when one walk out of his room it was like walking into a blast furnace (summer time in Phoenix). I am thin bone so I can take the heat as cold weather goes right through me, but Phoenix was something else.

The "snowbirds" from up North come to Phoenix in the winter time where the temperature is tolerable, but during the summer months, "let the Indians have it, I'm out of here".
 
Thanks all for the help (and humour) so far. I understand it until we solve for K and then I don't understand how you went about doing that.. I get stuck with 3/2u and 5/2...

:/

I'm trying to teach myself this whole course and it's super hard to do!
 
klooless said:
Thanks all for the help (and humour) so far. I understand it until we solve for K and then I don't understand how you went about doing that.. I get stuck with 3/2u and 5/2...

Don't follow where you are stuck? Where is u coming from??

:/

I'm trying to teach myself this whole course and it's super hard to do!
Click to expand...
 
I guess I was attempting to find k by substituting e^k as "u" to make it simpler. I don't know how to solve for k...
 
klooless said:
I guess I was attempting to find k by substituting e^k as "u" to make it simpler. I don't know how to solve for k...
Click to expand...

That's a good step - show us what you get after substitution - why are you stuck? Remember e[sup:sgpd38p7]2k[/sup:sgpd38p7] = u[sup:sgpd38p7]2[/sup:sgpd38p7]
 
So starting to solve for k:

9= [(6 e^2k)/(e^k - 1)] - [6/(e^k -1)] = (6e^2k - 6)/(e^k - 1) ----> 9/6 = (e^2k - 1)/(e^k - 1) , then u= e^k

9/6 = (u^2 - 1)/(u - 1) .....

This is where I am stuck :(
 
klooless said:
So starting to solve for k:

9= [(6 e^2k)/(e^k - 1)] - [6/(e^k -1)] = (6e^2k - 6)/(e^k - 1) ----> 9/6 = (e^2k - 1)/(e^k - 1) , then u= e^k

9/6 = (u^2 - 1)/(u - 1) = (u+1)(u-1)/(u-1)

3/2 = u + 1

u = 1/2

e^k = 1/2

k = ln(1/2) = - ln(2) ... now you are unstuck
.....

This is where I am stuck :(
Click to expand...
 
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