What is the Solution to Part B in Newton's Law of Cooling Problem?

[cbarker1]

Dear Everyone,

I need some help figuring part b in this problem.

In the book states, "According to Newton's Law of Cooling, if an object at temperature T is immersed in a medium having a constant temperature M, then the rate of change of T is proportional to the difference of temperature M-T. This gives the differential equation $\frac{dT}{dt}=k(M-T)$."

a) Solve the differential equations

b) A thermometer reading $100^{\circ}$F is placed in a medium having a constant temperature of $70^{\circ}$F. After 6 mins, the thermometer reads $80^{\circ}$F. What is the reading after 20 mins?

The Work for part a:

$\frac{dT}{dt}=k(M-T)$

$\frac{dT}{M-T}=k dt$

$\int{\frac{dT}{M-T}}=\int{kdt}$

Let $u=M-T$
$du=-dT$

$\int\frac{du}{u}=-k\int{dt}$

$\ln\left({\left| u \right|}\right)=-kt+C$

$e^{\ln\left({\left| u \right|}\right)}=e^{-kt+C}$

$\left|u \right|=e^{-kt}*e^{C}$

$u=\pm Ae^{-kt}$

$M-T=Be^{-kt}$

$T(t)=M-Be^{-kt}$

part b:
T(0)=100
T(6)=80
T(20)= ?
M=70 ?
I have trouble figuring out k and M.

Thanks for your help

Carter

 

[MarkFL]

I would choose to express the solution as:

\(\displaystyle T(t)=\left(T_0-M\right)e^{-kt}+M\)

where $T_0=100,\,M=70$ thus:

\(\displaystyle T(t)=\left(100-70\right)e^{-kt}+70=30e^{-kt}+70=10\left(3e^{-kt}+7\right)\)

Now we need to find the heat transfer coefficient $k$. We are told:

\(\displaystyle T(6)=80\)

Thus:

\(\displaystyle 10\left(3e^{-6k}+7\right)=80\)

Can you proceed to solve for $k$?

 

[cbarker1]

$k-\frac{1}{6}\ln\left({\frac{1}{3}}\right)=\frac{1}{6}\ln\left({3}\right)$

I got the final answer right.

Thanks for the help.