If I have a line of copper wire (lets say 1 meter long, 1mm thick) and one end is a flattened disk of copper about the size of a quarter, and I apply a lot of heat to it (I'm talking 800 Celsius) will the entire line be heated to the same degree? I mean what temperature will the unheated end be after, say, a minute? Can it too reach 800 degrees over time? how much time? And lets assume it's in a vacuum so no air removes any of the heat the copper holds.
[Physics] How effectively does heat flow through copper wire
temperaturethermodynamics
Related Solutions
The system will eventually reach thermal equilibrium so heat received by the gas is the same as heat lost by the container. So: $m c (T_f-T_i) + m c(T_f-T_i) = 0$. Hence you need to know the specific heat capacity of the materials, and the mass, as well as the initial temperatures.
The formula relating pressure, volume, amount of moles and temperature is: $PV = nRT$, where R is the gas constant (which depends on what units you use). From this you derive $P_1/T_1 = P_2/T_2$ so by inputting $P_1$, $T_1$, and $T_2$ you can calculate the final pressure.
EDIT: I do not believe you can easily calculate the time this takes.
With a constant volume for the container, the sum of the volume of the liquid and vapor is constant.
Let's start in an equilibrium position, temperature $T_1$. Volume of liquid is $V_l$ and vapor is $V_v$. Vapor pressure = $P_v$. Now we increase the temperature to $T_2$.
The first thing that happens is that the pressure of vapor increases to $P_v\frac{T_2}{T_1}$ before any further evaporation takes place (ideal gas law) and the temperature of the liquid also increases. The question is - do we expect more liquid to evaporate? In other words - is the increase in saturated vapor pressure faster than the increase in pressure with temperature due to the ideal gas law?
Now the August equation is a simple representation of the relationship between pressure and temperature, and takes the form
$$log_{10}P = - \frac{B}{T}$$ (this is really a different formulation of the Clausius-Clapeyron equation).
We can rewrite this as $$P = a e^{-b/T}$$ From this it follows that $$\frac{dP}{dT}=\frac{abe^{-b/T}}{T^2}=\frac{bP}{T^2}$$ Compare this to the ideal gas law $$PV = nRT\\ \frac{dP}{dT} = \frac{nR}{V} = \frac{P}{T}$$
The increase in vapor pressure will be greater than the increase in pressure due to the ideal gas law if
$$\frac{bP}{T^2} > \frac{P}{T}\\ b > T$$
So that leaves us the question - what is this factor $b$, and how does it relate to the stated fact that the specific volume is less than the critical specific volume?
Here I have to say - I am not sure. I do know that $b = \frac{\Delta H}{R}$, but in principle for a liquid with a very low enthalpy of evaporation, $b$ could be very small. It's been a long time since I did thermodynamics, and I am stuck on this very last part. What does this have to do with the critical specific volume?
I'm nonetheless posting this as an "answer", hoping that it will give somebody else the nudge needed to create a complete answer (or give me a hint so I can finish this myself...). It's something simple - but it's late here.
Anybody want to take a shot at finishing this?
Best Answer
Heat transport here obeys the heat diffusion equation $\partial_t T = \chi \partial^2_{xx} T$ where T(x,t) is the temperature and $\chi$ is the heat diffusivity. For copper $\chi \sim 10^{-4}$ m$^2$/s, according to http://en.wikipedia.org/wiki/Thermal_diffusivity. Neglecting heat losses (thermal radiation in vacuum) we can estimate that in time $\tau \sim L^2/\chi$ where $L\sim$1 m is the wire length, the temperature will be pretty well equilibrated along the wire. Putting the numbers in, $\tau \sim 1/10^{-4} = 10^{4}$ s $ \sim $ 1 hour. An exact solution (in the form of a series) can be easily obtained if radiation is neglected. In particular it will show that the temperature at the unheated end will reach 800 C asymptotically in time (but will get close to it in $\tau \sim L^2/\chi$). With radiation, the problem becomes nonlinear so a numerical solution will be needed. But I would expect for this relatively low temperature and assuming a realistic diameter wire the effect of radiation losses will be pretty small, probably it will produce a correction below 1% .