@JohnRennie hey

@psa hi :-)

trying to figure out (b)

any ideas?

Give me a moment to read it ...

for sure

If you've done (a) then can't you use this for (b)?

yep I've done (a) but

how exactly would I use that for (b)?

I assume you'll choose a sine wave for $T_1$ since any oscillation in temperature could be Fourier anaylysed into a sum of sine waves.

I used complex exponentials, but yeah the real part would be a cos

@psa what was your solution for (a)?

$$T_0 + \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dkdt'T_1(t')e^{i\alpha k t'}e^{-i\alpha kt}e^{-\sqrt{-ik}x}$$

that

I knew I wanted to use a Fourier transform in time, so I chose to force the time equation (after separating variables) to be $\dot g = -\alpha ikg$ by choosing the separation constant $-ik$ for the x-equation.

And if you now assume a specific value for $T_1(t)$ e.g. $Ae^{i\omega t}$, you can evaluate the integrals to get a solution in closed form. Yes?

for (b) I wanted something like the depth $x$ for which $T(x,t) = 0.1*T(0,t)$ but I'm not sure how to account for "daily" and "annual" surface variations

I'd guess it's going to come out something like:

huh no I actually didn't do that

$$ T(x,t) = T_0 + A e^{i\omega t} e^{-ax} $$

why is assuming $T_1$ of that form realistic in this example?

If you have a daily variation in temperature it can be Fourier analysed as a sum of sine/cosine waves with periods 24 hrs, 12 hrs, 8 hrs, and so on. Yes?

ah i see, yes

so we could choose $\omega = 2\pi/T$ where $T$ is the period of a day

?

Yes

And the leading term, i.e. a period of one day, will be the largest component.

So to a good approximation you can consider $T_1(t)$ to be just a sine wave with a period of 24 hours.

That's probably good enough to work out the 10% depth.

Note that it then asks about yearly variations, and you'll use the same tactic there.

hmm

In fact, since the heat equation is linear, the actual $T(x,t)$ will just be the sum of the solutions for the individual frequency components.

right

so we'd just need the value of $x$ such that $T(x,t)_{\omega = 2\pi/T} = 0.1*T(0,t)$

which should be pretty simple

I've just done a quick calculation, and if we assume $T(x,t) = T_0 + A e^{i\omega t} e^{-ax}$ then I find this is a solution with $i\omega = \alpha a^2$

So we get the depth variation to be $\Delta T(x) \propto e^{-x\sqrt{\omega/\alpha}}$

you just plugged $T_0 + Ae^{i\omega t}e^{-ax}$ back into the heat equation, I assume?

Yes

@JohnRennie did you take the change in temperature with respect to the surface?

i.e. $\Delta T = T(x) - T(0)$?

Yes. It's a lot more convenient that way.

right

then we just need where $\Delta T = 0.1$

Yes i.e. $e^{-x\sqrt{\omega/\alpha}} = 0.1$

man

I'll never have that kind of intuition!

Just wait 40 years :-)

haha

@JohnRennie I'm wondering about what this implies about geothermal heat pumps

@psa it tells you how deep you have to bury your heat collectors to make them immune from annual temperature variations.

ahhh

@JohnRennie why do you want the collector to not be affected by temperature variations on the surface?

I mean, wouldn't you want to account for that so that heat flows up in the winter and down in the summer?

I think in geothermal collectors the hot end is a lot hotter than the surface at all times. So it isn't a form of heat storage where you pump heat down in summer and retrieve it in winter. The heat flow is always upwards to the surface.

Oh, wait, I misread the question.

Yes, you want it to be a source in winter and a sink in summer.

But my comment still applies. Too near the surface and the ground will be cold in winter when you want to extract heat from it, and warm in summer when you're trying to pump heat into it.

So you want to go deep enough for the temperature to stay constant all year round. That way it's colder than the surface in summer and warmer than the surface in winter.

@JohnRennie Hi!

@YouKnowMe hi :-)

If electronegativity is not measurable, what's this?

@JohnRennie

Who said electronegativity is not measurable?

@JohnRennie toppr.com/ask/question/…

I guess it depends on what exactly you mean by measurable. The electronegativity can be calculated from measurable properties but I suppose it can't be directly measured.

@JohnRennie Measurable properties..?

Yes, you can measure bond polarisation for loads of different molecules and from that calculate electronegativities for the elements.

@JohnRennie So what should I write in exam?

Hmm, another of those situations where what matters is not the answer but what the exam wants ...

Safest option is to say it cannot be measured.

And it does indeed vary depending on the chemical environment of the atom. The figures in your chart are averages.

Thanks :-)

@JackRod Hello

hey

Why here n factor of h2s is not 8 ?

I can't see the picture

just wait it is getting load

my data is over so it will take some few minutes to load hope u can understand

No problem

Short size

do you know what is the net oxidation state of h2so4

it is +7

Of?

Sulphur?

No h2so4

Net charge us 0

and h2s is -1 it is very strong reducing agent

Ok

so what we see is H2SO4 + 8H+ + 8e- → H2S + 4H2O

I just balanced the electron

its n factor is 6

Yes but book?

ok?

Ok

I got it

Thanks

Hello @JohnRennie sir

@PrateekMourya hi :-)

Tomorrow is my chemistry test

So

No physics

Good luck! :-)

Meanwhile a short question

Do you have some examples of scientists

Who went through a normal schoollife

And did great research in college

Life

You mean made notable discoveries while they were still students?

I mean they lived a normal schoollife

But in college life they did research

Pretty much every scientist I can think of had a normal school life.

Including you?

Yes. I was top of my class, but I wasn't a genius at school. Very few scientists are.

Thanks

Today i was bothered that i haven't did any great for my dream

Since schoollife

Scientists usually only really start to do advanced things when they do their PhD.

So i was bit feared

I ended up studying a subject that I didn't even know existed when I was at school :-)

:)

Thanks for short great motivation

I suspect most scientists don't really know what is going to be their life's work until the final year of their degree.

XD

The best advice I can give is to study everything you can because you never know what is going to really interest you.

Well, i guess the best advice is to get a good rank in the JEE first :-)

Its like opening gate of scientific world

In india

If you get good rank

You can choose your favourite college

Else they will decide

Yes, though if you're really interested you will do well wherever you study.

I just stared some of your messages as they are so amazing

:-)