So if these hockey pucks absorb near 100% of the light, doesn't that mean they can't do that for very long due to the fact they would be heating up? The energy won't just disappear so it must be being converted to heat, unless I am totally off base here.
So if these hockey pucks absorb near 100% of the light, doesn't that mean they can't do that for very long due to the fact they would be heating up? The energy won't just disappear so it must be being converted to heat, unless I am totally off base here.
The power output of the lasers is probably of the order of 50 mW, so after an hour the material will have absorbed 180 Joules which isn't much.
@DeepThought Yes, that was my basic thought, low power for now but there must be an upper limit of power that this technology could handle.
180 Joules sounds like a lot, if you have 2 each 0.05 watt lasers, 100 milliwatts pumping for one hour, sounds like more like 6 joules isn't it? One Joule= one watt for one second? One hour at one tenth of a watt?
@sonhousesaid @DeepThought Yes, that was my basic thought, low power for now but there must be an upper limit of power that this technology could handle.
180 Joules sounds like a lot, if you have 2 each 0.05 watt lasers, 100 milliwatts pumping for one hour, sounds like more like 6 joules isn't it? One Joule= one watt for one second? One hour at one tenth of a watt?
3600 * 2 * 50.0E-3 = 3.6*100 = 360Joules, or twice my answer as I only calculated for one laser. According to Wikipedia [1] the specific heat capacity of steel is 466 J/Kg Kelvin. So an otherwise thermally isolated 100g steel target would increase in temperature by 7.72 Kelvin over an hour. Unless the target is very light or the experiment is in a vacuum I doubt the target heats up much.