heatthermodynamics
Suppose I have surface A in contact with surface B, if I apply Fourier's law of heat transfer, which $K$ should I use, $K_a$ or $K_b$?
Essentially asking whether the same block of material heats faster in 300 degree water or 300 degree air or the same.
John's answer is a good one, I just wanted to add some equations and addition thought. Let me start here:
Heating is really only significant when you get a shock wave i.e. above the speed of sound.
The question asks specifically about a $200^{\circ} C$ increase in temperature in the atmosphere. This qualifies as "significant" heating, and the hypothesis that this would only happen at supersonic speeds is valid, which I'll show here.
When something moves through a fluid, heating happens of both the object and the air. Trivially, the total net heating is $F d$, the drag force times the distance traveled. The problem is that we don't know what the breakdown is between the object and the air is. This dichotomy is rather odd, because consider that in steady-state movement all of the heating goes to the air. The object will heat up, and if it continues to move at the same speed (falling at terminal velocity for instance), it is cooled by the air the exact same amount it is heated by the air.
When considering the exact heating mechanisms, there is heating from boundary layer friction on the surface of the object and there are forms losses from eddies that ultimately are dissipated by viscous heating. After thinking about it, I must admit I think John's suggestion is the most compelling - that the compression of the air itself is what matters most. Since a $1 m$ ball in air is specified, this should be a fairly high Reynolds number, and the skin friction shouldn't matter quite as much as the heating due to stagnation on the leading edge.
Now, the exact amount of pressure increase at the stagnation point may not be exactly $1/2 \rho v^2$, but it's close to that. Detailed calculations for drag should give an accurate number, but I don't have those, so I'll use that expression. We have air, at $1 atm$, with the prior assumption the size of the sphere doesn't matter, I'll say that air ambient is at $293 K$, and the density is $1.3 kg/m^3$. We'll have to look at this as an adiabatic compression of a diatomic gas, giving:
$$\frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{\frac{\gamma-1}{\gamma}}$$
Diatomic gases have:
$$\gamma=\frac{7}{5}$$
Employ the stagnation pressure expression to get:
$$\frac{P_2}{P_1} = \frac{P1+\frac{1}{2} \rho v^2}{P1} = 1+\frac{1}{2} \rho v^2 / P1 $$
Put these together to get:
$$\frac{T_2}{T_1} = \left( 1+\frac{1}{2} \rho v^2 / P1 \right)^{2/7}$$
Now, our requirement is that $T2/T1\approx (293+200)/293 \approx 1.7$. I get this in the above expression by plugging in a velocity of about $2000 mph$. At that point, however, there might be more complicated physics due to the supersonic flow. To elaborate, the compression process at supersonic speeds might dissipate more energy than an ideal adiabatic compression. I'm not an expert in supersonic flow, and you can say the calculations here assumed subsonic flow, and the result illustrates that this is not a reasonable assumption.
addition:
The Concorde could fly at about Mach 2. The ambient temperature is much lower than room temperature, but the heatup compared to ambient was about $182 K$ for the skin and $153 K$ for the nose. This is interesting because it points to boundary layer skin friction playing a bigger role than I suspected, but that is also wrapped up in the physics of the sonic wavefront which I haven't particularly studied.
You have to ask yourself, what pressure is the nose at and what pressure is the skin at. The flow separates (going under or above the craft) at some point, and that should be the highest pressure, but maybe it's not the highest temperature, and I can't really explain why. We've pretty much reached the limit of the back-of-the-envelope calculations.
(note: I messed up the $\gamma$ value at first and then changed it after a comment. This caused the value to go from 1000 mph to 2000 mph. This is actually much more consistent with the Concorde example since it gets <200 K heating at Mach 2.)
Summarising the state of the art of heat transfer knowledge isn't easy, considering small libraries have been filled on this subject alone but I'll give it a shot.
Heat transfer occurs following three quite distinct mechanisms (or modes, if you prefer):
1. Radiative heat transfer:
Jim's answer to this question deals adequately with this so I don't have to.
2. Heat conduction:
When temperature gradients (generally $\nabla T$) exist in an object then by Fourier's law, heat conduction will strive to minimise these gradients and uniformise the temperature throughout the object.
3. Convection:
Convection combines radiative and conductive heat transfer with mass transport of heat: a domestic radiator (for instance) heats up the air surrounding it, which then rises because its density has been lowered. Convection can also be forced by means of ventilators that force the air (or a liquid) to flow over the hot surface.
So, to answer the OP's question, which model (equation) to choose?
Let's look at a hot sphere to focus attention (ignore the 'dent' due to poor drawing skills):
![Temperature gradients in a sphere]](https://i.stack.imgur.com/T63TO.png)
We're looking at a cut straight through the centre of a hot sphere (hotter than the surrounding temperature $T_{\infty}$).
The left hand side is a solid sphere and due to heat conduction the boundary of the sphere will be cooler than the core, as schematised by the temperature distribution curve $T(r,t)$, here at a specific time $t$.
The flow of heat inside the sphere is governed by Fourier's heat equation:
$$\frac{\partial T}{\partial t}=\frac{k}{\rho c_p}\nabla^2T$$
Where $\frac{k}{\rho c_p}=\kappa$ is the thermal diffusivity of the material.
Heat flow from the boundary of the sphere to the surrounding medium is then either through convection, radiation or a combination of both.
The right hand side is a hollow sphere filled with a liquid (or a fluid, more generally) that is constantly stirred. Due to this stirring there are no temperature gradients and thus no internal heat conduction:
$$\nabla^2T=0$$
Heat flow from the boundary of the sphere to the surrounding medium is then either through convection, radiation or a combination of both. Usually Newton's cooling law can be used to model that situation.
Choice of model:
The scientist/engineer will have to choose to model his real world system according to which of the two options, $\nabla^2T=0$ or $\nabla^2T\neq0$, best describes his system.
That choice will also be influenced by mathematical considerations: models that require use of Fourier's equation tend to be mathematically more demanding. Analytical solutions may not exist or be difficult to use. Some examples of using the Fourier equation for 1D heat transfer problems can be found in that link and illustrate the (relative) difficulty in obtaining analytical solutions. Numerical (computer) solutions may be preferred in many cases.
Best Answer
I'm going to add some mathematical detail to what akhmeteli has said.
Let's restrict the discussion to one dimension with coordinate $x$, then Fourier's law in differential form says $$ q(x) = -k(x) T'(x) $$ where $q(x)$ is the local heat flux, $k(x)$ is the conductivity, and $T(x)$ is the temperature gradient. Notice that Fourier's law show that that a given point, the derivative of the temperature is important, but derivatives of a function depend on the value of that function in a neighborhood of that point, not just the value of the function at that particular point. Therefore, the (not entirely explicit) answer to your question is that you need both $k$'s at a point where two materials with different $k$ are in contact. Now let's see the math.
If you are looking at a point $x_0$ at which two materials with different conductivities are joined, (say $k_a$ corresponds to $x<x_0$ and $k_b$ corresponds to $x>x_0$, then $k(x)$ has a jump discontinuity that can be written with the use of the Heaviside step function $\theta(x)$; $$ k(x) = (k_b-k_a)\theta(x-x_0) + k_a $$ Which results in the following differential equation: $$ q(x) = -[(k_b-k_a)\theta(x-x_0) + k_a] T'(x) $$ Which you can attempt to solve in a given case. For example, let's consider a steady-state system in which $q(x) = q_0$ is a constant and for which we want to determine the temperature gradient. Let's suppose that this system consists of metal bars joined at the point $x_0$ and whose endpoints are located at $x_0-L$ and $x_0+L$ respectively. Additionally, we assume that these other two endpoints are kept at a temperature $T_0$ In this case, the differential equation we would want to solve for $T(x)$ is $$ q_0 = -[(k_b-k_a)\theta(x-x_0) + k_a] T'(x) $$ with the boundary data $$ T(x_0-L) = T_0,\qquad T(x_0+L) = T_0 $$ The differential equation we want to solve can be rewritten as a set of two equations, one for $x<x_0$ and another for $x>x_0$; $$ q_0 = -k_a T_a'(x), \qquad q_0 = -k_b T_b'(x) $$ The general solutions are $$ T_a(x) = T_0-\frac{q_0}{k_a} [x-(x_0-L)], \qquad T_b(x) = T_0-\frac{q_0}{k_b} [x-(x_0+L)] $$ and the temperature everywhere except at $x=x_0$ can be written as $$ T(x) = (T_b(x) - T_a(x))\theta(x-x_0) + T_a(x) $$ In particular, notice that there is a jump discontinuity in the temperature at $x=x_0$; $$ T_b(x_0) -T_a(x_0) = q_0 L\left(\frac{1}{k_b}-\frac{1}{k_a}\right) $$ and this discontinuity depends on both $k$ values, not just the value on a particular side. Note further that if $k_a = k_b$, the the discontinuity disappears as you might intuitively expect!
Hope that helps! Let me know of any typos.
Cheers!