Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top heat a piece of steel so its glowing yellow (1100 C)? Assuming you had a cloudless day at a latitude of, say, San Francisco...

Basically I'm wondering if it is possible/feasible to be able to do basic metal working without a traditional forge, just using the power of the sun to heat the metal. So the diameter of the heated spot would have to be about 6" in order to heat a large enough area of the metal to work it...

I always thought you would need several huge pieces of equipment to do this, but just thought I'd ask if anyone here knew how to figure out it roughly...


share|cite|improve this question
up vote 1 down vote accepted

For steel, the specific heat would be $c_p=0.5 kJ/kg K$, with a density of $ \rho=7000 kg/m^3$. Suppose you want to increase the temperature bij say $\Delta T=1100K$ of a piece of size $V=(15cm)^3$

Then you would need a total energy of.

$$E=\rho c_p V \Delta T$$ Which gives you typically $E=10^7 J$

Now, the power of the sun on a bright day, would be of the order of $p=10^3 W/m^2$.

Assuming that

  • all the energy input is converted into heat
  • the mirror is perfectly aligned
  • no heat is lost during heating,
  • no melting, e.g. no latent heat

and your mirror had diameter $D$ and you let the process run for a time $t$, then

$$E=p \frac{\pi}{4}D^2 t $$

Then you will get, in approximation

$$D = \sqrt{\frac{E}{pt}}$$

So, suppose you are willing to wait for ten minutes, then the mirror diameter would be $D\approx 4m$. Considering we assumed an ideal system, this is only an order of magnitude assumption.

share|cite|improve this answer
One should also take into consideration whether the supplied power is sufficient to heat up the steel more while energy is lost due to thermal radiation. Assuming that it is a black body, we have $P=\sigma T^4A$ where $T=1500\textrm{ K}$, $\sigma=5.67\times 10^{−8}\textrm{ Wm}^{-2}\textrm{K}^{-4}$ and $A\approx 0.1\textrm{ m}^2$ which gives us $P=2.9\times10^4\textrm{ W}$. In other words, in order to keep it at such a temperature, we need $D \geq 6\textrm{ m}$. This is obviously a rather close call, as this power is absolutely required, while the power supplied is more of an upper bond. – Claudius Nov 20 '12 at 21:16
@Claudius, I assumed no heat losses, so the $4m$ is probably an absolute lower bound. But including it would demand solving a differential equation in time, as you reach a asymptotic temperature, didn't think it was necessary for an order of magnitude estimation. Only time $t$ is rather arbitrarily chosen in my estimation here. – Bernhard Nov 21 '12 at 7:01
If you still want to get the time $t$, you indeed have to solve a differential equation. However, if you ignore $t$ (i. e. $t \to \infty$) you can get an upper bound on the temperature reachable with a specific diametre (as given in my calculation above). Not including heat loss would effectively mean that if one waited long enough, any mirror would do. – Claudius Nov 21 '12 at 10:41
@Claudius You are absolutely right. – Bernhard Nov 21 '12 at 13:51

protected by Qmechanic Jan 7 '13 at 8:16

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.