For example I have a 1 watt laser and direct it to a sheet of metal (copper), if I were to direct it for say a time interval of 1 minute what would be the change in temperature? I can predict that it would depend on the size of the sheet as a larger sheet can dissipate more energy, the time exposed and the power of the laser, however I would appreciate someone walking me through.

# [Physics] Calculate the rise in temperature from directed radiation

energyhomework-and-exerciseslaserthermodynamics

## Best Answer

The answer depends on many factors, but here are the basic bits of physics that play:

The reflectivity of the copper is a particularly important one. If you have a well polished sheet of copper, the emissivity might be as low as 0.03 (reference) or, if it became oxidized from heating, as high as 0.78. That obviously has a HUGE impact on the tempeature dependent solution (but interestingly, if the final heat loss mechanism is only radiative, then it would not affect the steady state solution - only how long it would take to get there).

Let's first compute the steady state solution - it is a little bit easier than the transient, and it gives us some of the principles. I will assume a sheet of thickness $t$, focal spot of diameter $d$ with power $P$, emissivity of the surface $\epsilon$. That means that power absorbed is $\epsilon P$. The way this heat leaves the focal spot area is through conduction into the metal, and then through heat loss at the surface (radiation, convection). Radiative heat loss goes as $\epsilon \sigma_B T^4$ per unit area, while convection depends on mass flow of air (which in turn depends on orientation...). All quite complicated - easy to be off by factors of 2x and greater with "reasonable" assumptions.

But let's assume 1 minute with 1 W laser (as in your question), and emissivity of 0.03 (polished copper). Then the total heat absorbed by the copper is 1.8 J. The specific heat capacity of copper is roughly 0.4 J/g/C. If the sheet of copper is 1 mm thick, and circular with an area of 100 cm^2, then the total mass is approximately $\rho V = 8960\cdot 10^{-3-2} kg = 90 g$. This means that the heat capacity of the copper sheet is about 36 J/C, and that on average it will be heated by 0.05°C. That is assuming that thermal conductivity is sufficient to get the heating evenly across the entire sheet, which is probably false; but it does suggest that radiative and convective heat losses are likely quite small on this time scale.

So let's look at the thermal conductivity. Because the sheet is much thinner than it is wide, I am going to treat as a 2D sheet - as though all the heat is applied over a circle of diameter $d$. The lateral thermal conductivity of the sheet can be estimated from

$$k_{2D} = k_{3D}*t = 400 * 10^{-3} = 0.4 W/K$$

A given heat flux $\Phi$ at a radius $r$ will cause a thermal gradient

$$\frac{dT}{dr} = \frac{\Phi}{2\pi r k_{2D}}$$

so with the numbers given above, we would expect

$$\frac{dT}{dr} = \frac{0.03}{2\pi r \cdot 0.4} = \frac{0.012}{r}$$

Obviously, you need some boundary conditions. If we assumed for a moment that the edge was kept at room temperature, the thermal profile (integrating from the edge inwards) would be of the form

$$T(r) = T_0 - 0.012\cdot \log\frac{r}{r_{outer}}$$

This means that for steady state, with the edge clamped at room temperature we would expect the temperature rise in the middle to be only about 0.03°C. That suggests that thermal conductivity of the copper is really quite effective for this kind of heat load. If you make the sheet much thinner, and the focal spot smaller, the answer will obviously change.

As you start playing around with these assumptions, the simplifications that worked above might break down - and you quickly end up with a tricky diffusion equation. But sometimes, before you get all carried away with those, it's a good idea to estimate the magnitudes. In this case, the magnitude of the temperature rise is quite small (with the values I chose).