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
[Physics] Determining the temperature at the end of a rod when one end is heated
energythermodynamics
Related Solutions
Yes, your analysis is correct. Water in equilibrium with ice is at a temperature of 0 degrees C. The reason that the water doesn't spontaneously turn to ice has to do with the latent heat of fusion of water: in order for water to turn to ice at zero degrees C, you need to remove quite a lot of heat from it. In the case of water / ice, the latent heat is almost exactly 80 cal/g (334 J / g). Another way of thinking about this: one gram of ice at zero C can reduce the temperature of 1 gram of water from 80 C to zero C (or 4 gram from 20 to 0, etc). This is why you can add a few chunks of ice to a drink, and the whole drink will get cold without getting too diluted. Adding more ice than is needed helps speed up the process (because the surface area in contact with the liquid is larger) but you will end up with some solid ice at the end - and this in turn will keep the liquid at 0 C since any heat absorbed from the environment will quickly go into melting a bit more of the ice.
However, my hands seem to feel more comfortable being exposed to winter air than to snow
Air (still air) is a better insulator than snow. In fact the insulating properties of snow are due to the fact is has a lot of trapped air within it. The other component of snow, water, is not noted for being a good insulator.
The trouble with using air as an insulator is that it's hard to keep the air still. Even if there is no wind the convection currents caused by your hands/whatever/tub will cause the air to move. And as the air flows away it carries away heat and brings fresh cold air into contact with the object. Though your hands may feel warmer in still air then in snow, I bet they'd feel warmer in snow than in a 100mph blizzard.
If you look at how animals use snow burrows they typically rely on their fur (or blubber if they're seals) as the primary insulator. The temperature immediately outside them will typically be minus a few degrees C. Where the snow comes in is by insulating them from the -40°C gale that's howling outside their burrow.
In the case of your hot tub, assuming the polystyrene lid you mention can provide enough insulation that the temperature above it stays below 0°C then a layer of snow would add more insulation if the external temperature is well below zero. If the external temperature is only around zero then the layer of snow will achieve little.
Snow if obviously a useless insulator if the internal temperature is above zero. In that case the snow will melt and the latent heat of fusion required will cause a massive heat loss.
Best Answer
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 $L$.