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Lets say I leave 100g of copper outside under a convex lens that is 1m squared in surface area (I've been told that 1m2 of sunlight equals 1Kw). How would I figure out the temperature reached for a given amount of copper based on the heat it receives? What about the time it takes to reach that heat? I'm also concerned that a lot of the light would reflect off the copper and not become heat. What's a way around that? Thanks

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You could paint the copper black :)

In general, what you need is the reflectivity (or absorptivity) to figure out how much radiation actually makes it into the material (some fraction of incident light energy). Then, you need the emissivity, which tells you how well the material radiates as compared to a black body. Finally, you need the thermal diffusivity and solve the heat equation. The initial conditions are whatever temperature the material started at, and the boundary conditions are given by the heat flux through the surface that you calculated in the previous part. Of course, this ignores any interaction with surrounding air.

As the object gets hotter, it radiates more and more, so that at a certain point it is radiating as much energy as it is absorbing, and the temperature then stays steady.

The above is pretty messy, and you can make a few approximations, since you are dealing with such a small amount of metal and such a large lens. The specific heat capacity of copper is $0.385\,J/(gK)$. Throw in an absorptivity of $0.5$ (taken from here). Assume your optical setup is $10\%$ efficient, so that the sample absorbs $100\,J/s$. This shows us that initially your sample of copper will rise in temperature by about $1K$ a second (or $6K/s$ for a more generous $50\%$ efficient optical setup). Once you start taking other effects into account, the heating rate will be less than this, and will slow down as you reach the maximum temperature.

I forgot to mention that you can't heat the sample beyond about $5500\,K$ (which is the temperature of a black body that best fits the spectrum we receive at sea level). Of course, you need to know what you're doing to get anywhere close to this temperature.

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