How to calculate rate of water heating with pulsed near infrared laser? I'm designing something that requires me to heat a small volume $( <10 ml)$ of water with a $2100 nm$ pulsed near-infrared laser with settings between $1 J$ and $1.5 J$ and $5-15 Hz$. I'm heating the water in the range of $20-70$ degrees C. I thought that it would be fairly easy to calculate the rate of heating as below:
$$Q = mc\Delta T$$
$$\Delta T = \frac{Q}{mc}$$
$$Q = Eft$$
Therefore, 
$$\Delta T = \frac{Eft}{mc}$$
$$or$$
$$dT = \left(\frac{Ef}{mc}\right)dt$$
Experimentally, all of my heating curves are linear but all of my experimental heating rates are $0.6\left(\frac{Ef}{mc}\right)dt$ almost perfectly. Am I stupid and missing something simple up? Is there a coefficient specific to pulsed laser heating that would give me $0.6$ somewhere? I'm assuming that all of the energy from the laser is absorbed by the water because of the low penetration of infrared through water and my distance of $\approx 2 cm$.
 A: No, there is no specific effect related to heating with pulses compared with any other heating method. Your calculation should give the correct result. You should treat this as a systematic error and look for a physical explanation.
Perhaps your assumption of perfect absorption is incorrect? Absorption is not affected by pulse repetition rate, and probably is only very weakly dependent on temperature because decrease in density - ie the "concentration" of absorbers - is small.
The 1st graph in the wikipedia article on Absorption of EMR by water gives an absorption coefficient of about $\alpha \approx 2000 m^{-1}$ at $2.1 \mu m$. Your absorption length is $x=0.02m$ so over this length you should expect transmission of about $e^{-\alpha x} = e^{-40} \approx 4\times 10^{-18}$. This confirms your assumption that almost all the energy from the beam should be absorbed, so there must be some other explanation(s) for the energy loss.
Reflection at the air/glass front face of the heating cell would only account for about 3.5% of the loss. I am assuming the cell is made of glass.
Have you included in $mc$ the heat capacity of the cell which holds the water? (I presume not, since you take c=4.2.) And also of the thermometer?
If the time taken to heat the water is "long", and the apparatus is not thermally insulated, losses due to conduction/convection/radiation could be significant.
