Metal temperature change I have a pipe that’s $70^{\circ}\text{F}$, in a constant room temperature of $80^{\circ}\text{F}$.
I would like an equation to solve for pipe temperature ($\text{F}$) after X amount of time. 
Material: Iron
Height: $100~\mathrm{cm}$
Diameter: $3.81 ~\mathrm{cm}$
Based on this equation, using Newton’s Law of Cooling, I will determine the $k$ (rate of change in temperature/time).
NOTE: 
I understand that I’m not accounting for all influences.  Nevertheless, using the provided facts I would like the best solution.
 A: The simplest treatment of this kind of problem for an object of mass $m$ and specific heat capacity $c$ is by means of Newton's cooling law, for convective cooling only:
$$-\frac{dQ}{dt}=hA(T-T_{\infty})$$
Where $h$ is the heat transfer coefficient, $A$ the total surface area of the object and $T_{\infty}$ the ambient temperature.
Also:
$$dQ=mcdT$$
So:
$$-mc\frac{dT}{dt}=hA(T-T_{\infty})$$
Integrated between $T_1$ and $T_2$, this yields:
$$\Large{T_2=T_{\infty}+(T_1-T_{\infty})e^{-\alpha t}}$$
Where:
$$\alpha = \frac{hA}{mc}$$
$T_1$ is the initial temperature (at $t=0$) and $T_2$ is the temperature at time $t$.
The rate of change in temperature/time is of course not linear and given by:
$$\frac{dT}{dt}=-\alpha (T_1-T_{\infty})e^{-\alpha t}$$
If needed, the following logarithmic plot allows to determine $\alpha$ experimentally:
$$\ln\Big(\frac{T_2-T_{\infty}}{T_1-T_{\infty}}\Big)=-\alpha t$$
Note that this simple model is based on some assumptions:


*

*Heat loss by convection only.

*Temperature is uniform throughout the object's domain (no temperature gradients due to conductive lags).

*$m, c, h, A$ are temperature invariant.



Edit:
To answer OP's question in the comments: $T_2$ after $2\:\text{min}=120\:\mathrm{s}$, $T_1=70^{\circ}\text{F}$ and $T_{\infty}=80^{\circ}\text{F}$?
We'll neglect the iron shell and pretend the pipe is all water.
First calculate:
$$\alpha = \frac{hA}{mc}$$
$h \approx 20\:\mathrm{Wm^{-2}K^{-1}}$
$A=\pi D L=0.12\:\mathrm{m^2}$
$m=\rho V=\rho \frac{\pi D^2}{4}L=1.14\:\mathrm{kg}$
$c_{water}=42000\:\mathrm{Jkg^{-1}K^{-1}}$
$$\implies \alpha=0.0005\:\mathrm{s^{-1}}$$
$$T_2=80+(70-80)e^{-0.0005 \times 120}\approx 70.6^{\circ}\text{F}$$
Or more generally:
$$T_2(t)=80-10e^{-0.0005 \times t}$$
Alternatively to find the time $t$ to reach a specified temperature $T_2(t)$, use:
$$t=\frac{1}{\alpha} \times \ln\Big(\frac{T_{\infty}-T_1}{T_{\infty}-T_2(t)}\Big)$$
But in that case all temperatures need to be converted to Kelvin first.
For example to take the pipe to $75^{\circ}\text{F}$ in the above conditions would take about $135\:\mathrm{s}$:
$$t=2000\times\ln\frac{299.8-294.3}{299.8-297.0}=1350\:\mathrm{s}=22.5\:\mathrm{min}$$
A: In my thinking that you may use following equation
$\frac{dT}{dt}\propto-(T-T_0)$
$\frac{1}{T-T_0}dT=-kdt$
$T-T_0=Ce^{-kt}$
from graphical method you may estimate the value of k.
