I found some NIST Webbook data that indicates an absorption spectra around 423Mhz for methane. I believe that means that if I had a bottle of methane which I illuminated with microwaves at 423Mhz, I'd get an ongoing conversion of CH4 -> CH3 + H- then recombination back to CH4. Do I read that properly?

Also, Hi to The Periodic Table. I'm new.

@Green No, it's a rotational–vibrational transition.

@Loong So instead of removing a hydrogen, I would just end up heating the CH4?

@Loong Hi! I see you lurking in WB but don't remember talking to you before. Thanks for your help.

@Green To break a bond, you need much higher energies, because these involve transitions between electronic states, which are much further apart in energy than just rotational or vibrational states. Some bonds can be broken with visible light. For others you need UV. I don't know if it is possible to break one C–H bond in methane by itself, but if you could, the energy required would be in at least the UV range.

(rotational transitions = typically around microwave; vibrational transitions = typically around IR)

I've got a question about thermochemistry. Is this the right time, or am I interrupting?

Eh, it's been over a third an hour. I'll go ahead:

I have the heat capacity of a container, the mass of water in it, the mass of iron dropped into it, and initial temperatures of the water and the iron. Since I know the substances, I know the specific heats.

How do I find the final temperature?

@user10535 I assume you start with cold water and hot iron?

Conservation of energy.

Yes, hot iron, cold water

So the iron will heat the water, and the water will cool the iron. And you know at which point this exchange will end.

@orthocresol Look at you, asking a good question and providing a good answer to it :)

No, I don't

I don't know what the final temperature is

But you know that the final temperature of the iron = the final temperature of the water

I have the final temp of neither

I know how to get the final temp when I don't have to account for the container

@Zhe Ah, it was kind of random. I'd prefer slightly more interesting questions than that, but because we deleted a similar (but poorly written and answered) question a while ago in Spring Cleaning, I wanted to write a proper Q+A to replace it.

But what about when I need to consider the heat capacity of the container?

That's when it becomes an engineering problem and no longer a chemistry problem.

Then you will also heat the container and not just the water.

@orthocresol Thank you, that's very helpful.

If the container has the same initial temperature as the water, you can combine water and container.

Do I multiply its specific heat by 4.184 J/C?

@user10535 At some point, you need to worry about heat loss to the environment. Then it becomes a calculus problem.

It's a step above an introductory chem course

@Zhe Yes, calculate how much a better calorimeter costs.

Now, is it that the question is poorly designed?

@Loong :D

Turns out for simple stuff a double styrofoam cup loses very little heat and is pretty good

@user10535 Add the heat capacity of the container and the heat capacity of the water.

Oh

I never thought of that

Thank you

@orthocresol Unable to figure out how to superimpose them ...

Build a model

Ok let me try

By the way, all six vertices in the octahedron are equivalent. There isn't any "axial" or "equatorial" in an octahedral structure, they are all the same.

So there's no difference between an "equatorial"+"equatorial" cis isomer and an "equatorial"+"axial" cis isomer.

@orthocresol I dont have any atom with 6 holes :(

@Abcd chemtube3d.com/SymFeCl4Br2cis.htm

trans: chemtube3d.com/SymFeCl4Br2trans.htm

That might help.

@orthocresol Understood, thanks. Will delete my question now.