Chemistry brain teasers! Noob Competition!

Quote: Originally posted by aga

They're do-able, but f*king tricky tho. Mostly it is just following the Rules and applying Logic and some maths, so i guess knowing the Rules, Some Logic & Maths is kinda vital. Following those Rules and not jumping to easy-path conclusions is also pretty useful ! If we're Lucky (maybe if we ask nicely enough, offer money, sexual services, drugs etc) bloggers might show us all the Answers with a step-by-step walk-through showing how they can/should be worked out. @<b>NEMO-Chemistry</b> & <b>Neme</b> Which bits are the really hard parts for you ? Stoichimetry ? (aka Balancing Equations) Algebra ? (aka Squigly Maths-With-No-Numbers) Molarity ? (aka how much stuff in that solvent) If you shout up, you could get some help. For me, it's just thinking clearly that is quite hard ! [Edited on 11-9-2016 by aga]

I guess its perspective, a few short weeks ago i couldnt balance an equation, now i can do a little bit more than that.

So maybe its simply a question of keep studying and not let it phase you, i am working on the questions in sequence, so its likely things i have no idea of now will become clearer as i progress.

Maths is my main sticking point, but even that is starting to improve slowly. I think it was more of a knee jerk reaction reading the question and comparing it to the first one.

I have a clear week from Tuesday (school work wise), as i am almost upto date with my course work. This will give me more time to study this.

I will shout when i get totally stuck , i havnt finished the others yet as my course work needed getting upto date.

I guess there is little point in asking questions that dont stretch you anyway, no one learns anything of value if they already know the answer!

As it stands (a few more participants need to get in their answers on #7 and #8 but there's plenty time) the leader board is tight, with so many good answers.

Depending on the incoming answers a tiebreak question may be needed in 10 days or so. Maybe something with Calculus Lite, mwwhahaha...

[Edited on 11-9-2016 by blogfast25]

And as the Devil makes idle hands, here's a non-chemistry, Non-Competition question, just for fun.



A square metal plate has one side held at 100 C and the three other sides held at 0 C. Top and bottom of the plate are perfectly insulated (no heat losses of any kind).

Question:

After thermal equilibrium has been achieved, what is the temperature at the centre of the square, T_c?

[Edited on 11-9-2016 by blogfast25]

Quote: Originally posted by aga

$$p_{ain} = \int_0^{\infty} { \frac {aga}{calculusrack}} $$

Your integral lacks a differential.

Also, you have no idea how light "Lite" can become, in the limit!

[Edited on 11-9-2016 by blogfast25]

Quote: Originally posted by aga

Should random answers go via U2U ?

Yes, please. Forgot to mention that.

Quote: Originally posted by NEMO-Chemistry

Shit my answers must be crap, i didnt get feedback lol. How long we got? I will get these done this week.

Quote: Originally posted by NEMO-Chemistry

Ok i might have misunderstood!! I have been working on them and thought we were to submit them in one go?? When is the closing date and do i submit all together or as they are done??

Quote: Originally posted by ficolas

#3 too? yay missef another deadline

Non-competition question solution:



The question what is the temperature at the centre of the disc? was answered correctly as "25 C" by aga and Neme. Neither provided much rationale for their answer. So here's proof!

As stated in the original problem there are no radiative or convective losses, so the temperature distribution is described entirely by Fourier's heat equation, here in 2 dimensions and Cartesian coordinates:

$$\frac{\partial u}{\partial t}=k\Bigg(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}\Bigg)$$
u is temperature, k is the heat conductivity of the disc's material. Or using the usual shorthand:
$$u_t=k(u_{xx}+u_{yy})$$
Because we're in steady state, so u_t=0:
$$u_t=0 \implies u_{xx}+u_{yy}=0$$
Which is essentially Laplace's equation. Now we need to add the boundary conditions:
$$u(0,y)=0,u(L,y)=0$$
$$u(x,0)=0,u(x,L)=u_0$$
(Here u₀=100 C)

We can solve this by separation of variables from the following Ansatz:
$$u(x,y)=X(x)Y(y)$$
The solution is fairly straightforward and the process is described here. This gives us:

$$u(x,y)=\displaystyle \sum_{n=1}^{+\infty}A_n\sinh\Big(\frac{n\pi y}{L}\Big)\sin\Big(\frac{n\pi x}{L}\Big)$$
$$n=1,2,3,...$$
A_n is determined from one of the boundary conditions:
$$u(x,L)=u_0=\displaystyle \sum_{n=1}^{+\infty}A_n\sinh(n\pi)\sin\Big(\frac{n\pi x}{L}\Big)$$
Using Fourier series analysis we get:
$$A_n=\frac{2u_0}{n\pi\sinh(n\pi)}\big(1-(-1)^n\big)$$
So that T_c is given by:
$$T_c=u(L/2,L/2)=\frac{2u_0}{\pi}\displaystyle \sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)\cosh ((k+1/2)\pi)}$$
With:
$$k=0,1,2,3,...$$
Sums of infinite series can be a little scary but we can evaluate it in two ways:

1. Numerically:

Just the first term (k=0) gives:

$$T_c=0.2537 u_0$$

Then adding a few terms (k=1, k=2, etc) immediately shows the sum converges on:

$$T_c=0.25 u_0$$

2. Analytically:

Various databases of analytically determined infinite sums exist and this one shows that:

$$\displaystyle \sum_{k=0}^{+\infty}\frac{(-1)^k}{(2k+1)\cosh ((k+1/2)\pi)}=\frac{\pi}{8}$$

So that:

$$T_c=\frac{2u_0}{\pi}\times \frac{\pi}{8}=\frac{u_0}{4}=25 C$$

Plots:

For L=1, first 5 terms:





The squiggly edge is due to having used only 5 terms, using more terms smooths that edge more and more.


[Edited on 23-9-2016 by blogfast25]

End of current Noob competition: Friday 30 September, midnight UTC.

No more questions will be posted before that date.

Thanks all for playing!

[Edited on 27-9-2016 by blogfast25]

Thanks

Quote: Originally posted by CharlieA

When I taught chemistry for a few years, the part I liked least was grading papers.