Determining the temperature at the end of a rod when one end is heated If I have a rod of some material and submerge one end in hot water then leave the other side exposed to room temperature air. How would I go about calculating the temperature of the end of the rod that is exposed to the open air? It's alright if it requires two or three equations and some problem solving to figure out I'd just like some mathematical way to solve this if there is one
 A: We assume the rod to be thin (compared to its length) and uniform in shape and composition because that way the problem can be treated as a one-dimensional ($x$) problem.
The starting point is the (1D) non-homogeneous Fourier heat equation:
$$u_t=ku_{xx}+Q(x,t),$$
where $u$ is the spatial and $t$ the time distribution of temperature:
$$u(x,t)$$
and $Q(t,x)$ a heat loss function representing convection loss of the rod to the environment.
You are only interested in the steady state solution, so:
$$u_t=0$$
With some relevant substitutions we then obtain:
$$\frac{\mathbf{d^2}T}{\mathbf{d}x^2}-\frac{Ph}{Ak}(T-T_{\infty})=0,\tag{1}$$
where $T$ is the temperature of the rod, $T_{\infty}$ the surrounding air temperature (assumed constant), $P$ the perimeter of the rod, $A$ its cross-section, $h$ the convection heat transfer coefficient and $k$ the heat conductivity of the rod's uniform material.
Fistly, an easy substitution simplifies things a bit:
$$\tau=T-T_{\infty}\tag{2}$$
$$\mathbf{d}\tau=\mathbf{d}T\implies \mathbf{d^2}\tau=\mathbf{d^2}T$$
Insert into $(1)$ to get:
$$\frac{\mathbf{d^2}\tau}{\mathbf{d}x^2}-\frac{Ph}{Ak}\tau=0\tag{3}$$
This is a straightforward linear, second order, homogeneous differential equation. But we need two boundary conditions (BC) to obtain a solution.
Firstly, the value of $\tau$ at $x=0$, the heated end:
$$\tau(0)=\tau_0$$
For the second BC, we have some choices but a common one is to assume the end of the rod ($x=L$) doesn't conduct heat to the environment, which mathematically means:
$$\Big(\frac{\mathbf{d}\tau}{\mathbf{d}x}\Big)_{x=L}=0$$
Solving $(3)$ with the stated BCs and back-substituting with $(2)$ we obtain the temperature profile in function of $x$, in the following form:
$$\frac{T(x)-T_{\infty}}{T_0-T_{\infty}}=\frac{\cosh\big[\big(1-\frac{x}{L}\big)mL\big]}{\cosh(mL)},\tag{4}$$
where:
$$m=\sqrt{\frac{Ph}{Ak}}$$
and $T_0=T(0)$.
Setting $x=L$ in $(4)$ then allows to compute the temperature at the cool end of the rod, $T(L)$.

Edit: numerical example based on OP's numbers in comment section.
Based on $(4)$, for $x=L$, then:
$$\cosh\big[\big(1-\frac{x}{L}\big)mL\big]=\cosh(0)=1$$
Then:
$$T(L)=T_{\infty}+\frac{T_0-T_{\infty}}{\cosh(mL)}$$
OP provided the following data:
$A=0.0078\:\mathrm{m^2}, P=0.314\:\mathrm{m}, h=25\:\mathrm{W/m^2K}, k=50\:\mathrm{W/mK}, L=2\:\mathrm{m}, T_0=100\:\mathrm{C}, T_{\infty}=60\:\mathrm{C}$
This gives a value of:
$$m=4.49\:\mathrm{m^{-1}}$$
And:
$$T(L)=60+\frac{100-60}{\cosh(4.49\times 2)}=60.01\:\mathrm{C}$$
So for a length of $2\:\mathrm{m}$ the other end of the rod is basically at $T_{\infty}$. Play around with other values of $L$.
