How do we calculate the temperature of a laser-heated substance?

[Pericle]

Homework Statement

Knowing that laser beam before going through a cell with a 100µm layer of syrup has a power of 100mW and after going through a cell has 40mW, calculate what is the temperature of a syrup in a point where laser beam went through it?
Temperature of syrup before being heated is 25°C
Radius of the beam is r=0.35cm
Density of a syrup is d=1.08g/l

Relevant Equations

𝑃_𝑏/𝑃_𝑎
Q=mcT
P=πr[SUP]2[/SUP]
M=P*H*d

P_b/P_a =0.4
I have calculated mass of the syrup which is 0.42g
We know that the power of a laser before is 100mW and after is 40mW, that means that the energy absorbed by the syrup is equal to 0.6J/s.
I'm not really sure what to do here, the substance through which laser is going is Maple syrup, but I don't know any properties of it and I guess I can only make an assumption about them.

Do i just now only need specific heat?

How do i deal with heat that radiated away?

 

[kuruman]

If you know that the energy absorbed by the syrup is 0.6 J/s how many Joules are deposited by the beam when the syrup is zapped by the laser? (Hint: How long does it take the beam to go through the layer?) That number is your Q in the equation. Yes, you will need the specific heat. You can also assume that the process happens fast enough so that you don't have to worry about heat being radiated away.

 

[BvU]

Continuous laser ? Exposure time ?

 

[Pericle]

It is a continuous exposure.

I have also made a mistake, because 0.06J is absorbed only.

I got the mass wrong i think. It is supposed to equal 3.85•10^-7
Then the change in temperature is about 37K which actually makes sense

It's 0.06/(3.85•10^-7•4200)

 

[sojsail]

You said in your last post

0.06J is absorbed

. Did you intentionally change from the

J/s

units you used originally?

If it is a continuous exposure, the temperature will continue increasing until it radiates away energy at the same rate the laser is delivering energy.