Converting Electromagnetic Energy To Temperature

1. Mar 13, 2016

marmanq

1. The problem statement, all variables and given/known data
For a home work assignment, I have to build a theoretical device which uses a laser to boil water. But I have no idea how to figure out how long it would take the device to transfer enough heat to boil the water. The laser diode has a wavelength of 445nm. We are supposed to assume c=3.0*10^8.

2. Relevant equations
How would I get from E=hf, to ΔT for water?
Should I be using the Stefan-Boltzmann Law?

Any help would be appreciated, thanks!

2. Mar 13, 2016

drvrm

http://www.surface-tec.com/pldlaserheater.php

3. Mar 14, 2016

Buzz Bloom

Hi @marmanq:

What seems to be missing from the problem statement is the power of the laser in watts. The Stefan-Boltzmann Law is not relevant since that applies to thermal radiation, which a laser is not.

Hope t his helps.

Regards,.
Buzz

4. Mar 14, 2016

haruspex

How does the energy actually get into the water molecules?

5. Mar 15, 2016

Buzz Bloom

Hi haruspex:

The 445 nm wavelength corresponds to a wave number of ~22,500 cm-1. This is in the visible range and corresponds to violet. The following article shows that there is an absorption range of 3800-2800cm-1, corresponding to 2532-3571 nm. My guess is that water is transparent to 445 nm light and the laser would not heat it at all.

However, an IR laser might heat the water as follows.

A photon hits a molecule and based on the absorption coefficient of water for the frequency of the photon, there is a corresponding probability that it will be absorbed. If it is not absorbed, it will be scattered in a new direction. If the random walk path of the not absorbed photon is long enough before it leaves the water, there will be a high probability of its being absorbed causing it to enter an excited state. There is also an exponential probability distribution with respect to time that the molecule will spontaneously emit a photon of a similar wavelength and return to its former lower energy state. The mean of this distribution is the average time before such a emission will occur. This average time is related to the the Einstein coefficient for the molecule and the wavelength.
For water vapor, the molecules are generally far enough apart for this emission to almost always occur before the excited molecule will interact with another molecule. For liquid water (or ice) the excited molecule will most likely interact with another molecule before emitting a photon, and its excited state energy will become additional kinetic energy. The increase in the average kinetic energy of the water molecules would corresponds to an increase in temperature.

Regards,
Buzz

6. Mar 15, 2016

haruspex

Yes, that was the point of my question, but it was directed to marmanq. It has implications for the design of the device.

7. Mar 16, 2016

Buzz Bloom

Hi haruspex:

Sorry I misunderstood the pedagogical point of your question. It was too subtle for me.

Regards,
Buzz