So, say I have some mild steel. The steel is 250F. I take it away from the heat and place it in ambient temperature, $T_a$, now I'd like to know how long it takes to get from 250F to some arbitrary temperature, $T_{a'}$. I think for mild steel the heat transmission coefficient is 7.9 $\frac{W}{m^2K}$… Not sure how to set this up. Not homework. Actual real world problem in my life. I'd like to see how you got the answer, as I'd like to use it over and over again for different temperatures.
Edit:
Ok, so my initial conditions are $T(0s)=102F, T(270s)=122F$, and I'm working with $\frac{dT}{dt}=-k(T-T_a)$. Where $T_a=250F$ is the ambient temperature. Now in my new test case I'm heating a block of steel from ambient to 250F (assuming homogeneous dissipation and constant temperature regulation…). So, one thing to another we get this:
$\frac{1}{T-T_a}\textbf{dT}=-k\textbf{dt}$
$\int\frac{1}{T-T_a}\textbf{dT}=-\int k\textbf{dt}$
$\ln{|T-T_a|}=-kt+c$
$T-T_a=e^{-kt}e^{c}$
${
T(t)=
\begin{cases}
T_{>}(t)=Ce^{-kt}+T_a, T>T_a \\
T_{<}(t)=T_a-Ce^{-kt}, T<T_a \checkmark
\end{cases}
}$
So, for $T_{<}(0)$ we have
$102=250-Ce^{-0k}$
$148=C$
Ok, so then $T_{<}(270)=122$
$122=250-148e^{-270k}$
$122=250-148e^{-270k}$
$\frac{32}{37}=e^{-270k}$
$-\frac{\ln\frac{32}{37}}{270}=k$???
So I guess that makes
$T_{<}(t)=250-148e^{\frac{\ln\frac{32}{37}}{270}t}$
But… If I wanted to see how long it takes to get to 250F I get no solution? What's happening here?
Note: $T(0)$ isn't the true zero. It's staggered from an arbitrary time after heating has actually begun. The steel block starts at ambient 50F, and picked a random time after it got to 102, called that 0, let 270 seconds pass and marked the second time at 122.
Best Answer
This is the approach I took. From Newtons Law of Heating :) $$\frac{\mathrm{d}}{\mathrm{d}t}T(t)=k\left[T_a-T(t)\right]$$ $$\frac{1}{T_a-T(t)}\frac{\mathrm{d}}{\mathrm{d}t}T(t)=k$$ $$\ln\left|T_a-T(t)\right|=\ln\left(e^{kt+c}\right)=\ln\left(M{e^{kt}}\right)$$
SInce $T_a-T\gt 0$ $$\frac{1}{T_a-T(t)}=M{e^{kt}}$$
Using the values given $T_a=250$, $T(0)=102$: $$\frac{1}{250-102}=M{e^{k0}}\Rightarrow M=\frac{1}{148}$$
Using the values given $T_a=250$, $T(270)=122$: $$\frac{1}{250-122}=\frac{1}{128}=\frac{1}{148}{e^{k(270)}} \Rightarrow \frac{148}{128}=\frac{37}{32}=e^{k(270)}$$ Therefore: $$k=\frac{1}{270}\ln\left(\frac{37}{32}\right)$$
Putting it all together: $$250-T(t)=\frac{1}{e^{\frac{1}{270}\ln\left(\frac{37}{32}\right)t}}$$