writing equation for integration regarding rate of change

[poppy]

Hi

An object with temperature 26 degrees Celsius is placed in water with constant temperature of 90 degrees Celsius. If the temperature of the object rises to 70 degrees Celsius in five minutes, what will be the temperature after 10 minutes?

I thought of using Newtons law of cooling d(temp)/d(t) = -k(constant of proportionality) (temp(object)-temp(surrounding medium)) but have read on the internet that it only describes transfer of heat from object to water not the other way around.

Any ideas would be appreciated :wink:

 

[galactus]

Your instincts were correct. It is a Newton's Cooling Law problem.

It works both ways.

\(\displaystyle \frac{dT}{dt}=k(T-90)\)

Separate variables, integrate and all that. We get:

\(\displaystyle T=90+Ke^{kt}\)

Note, that T(0)=26 and T(5)=70.

You can use these two initial conditions to find k and K.

\(\displaystyle 26=90+Ke^{k(0)}\)

\(\displaystyle 70=90+Ke^{k(5)}\)

Then, you have an equation you can just plug t=10 into to find the temperature, T

 

[poppy]

Hi, that makes sense , however i was wondering where the second K comes from and where the constant of integration goes...

My integration gave me

dT=k(T-90) dt
1/(T-90) dT=k dt
ln|T-90|=kt+C
T-90=e(kt)+C
T=90+e(kt)+C

Thanks

 

[galactus]

Well, when we integrate we get:

\(\displaystyle ln|T-90|=kt+C\)**

Take e to both sides:

\(\displaystyle T-90=e^{kt+C}\)

\(\displaystyle T-90=e^{kt}\cdot e^{C}\)

Now, \(\displaystyle e^{C}\) is a constant we can rename K:

\(\displaystyle T=90+Ke^{kt}\)

This is what is commonly done.

**I forgot the absolute values in the beginning. I should have written 90-T. No big deal. Get same solution either way.

 

[BigGlenntheHeavy]

\(\displaystyle Newton's \ Law \ of \ Cooling: \ \frac{dT(t)}{dt} \ = \ k[T(t)-L]\)

\(\displaystyle Given: \ L \ = \ 90, \ T(0) \ = \ 26, \ T(5) \ = \ 70, \ Find \ T(10).\)

\(\displaystyle \frac{dT(t)}{dt} \ = \ k[T(t)-90], \ \implies \ \int\frac{dT(t)}{T(t)-90} \ = \ \int kdt\)

\(\displaystyle \implies \ ln|T(t)-90| \ = \ kt+C, \ \implies \ |T(t)-90| \ = \ (e^{C})(e^{kt})\)

\(\displaystyle \implies \ T(t)-90 \ = \ \pm(e^{C})(e^{kt}), \ = \ Ae^{kt}, \ A \ = \ \pm e^{C}\)

\(\displaystyle Hence, \ T(t) \ = \ Ae^{kt}+90, \ T(0) \ = \ 26 \ \implies \ A \ = \ -64\)

\(\displaystyle Ergo, \ T(t) \ = \ -64e^{kt}+90, \ T(5) \ = \ 70, \ \implies \ T(t) \ = \ -64(\frac{5}{16})^{t/5}+90\)

\(\displaystyle Therefore, \ T(10) \ = \ 83.75^{o} C, \ Note: \ \lim_{t\to\infty} T(t) \ = \ 90 \ as \ it \ should \ be.\)

 

[poppy]

Thanks for your time

I did manage to work it out on my own though
and it was the same as yours, so it must be right