No idea of how to really start this question

  • Thread starter Thread starter crysix
  • Start date Start date
  • Tags Tags
    Idea No idea
crysix
Messages
2
Reaction score
0
The average rate at which energy is conducted outward through the ground surface in a certain region is 63.7 mW/m2, and the average thermal conductivity of the near-surface rocks is 3.96 W/m·K. Assuming a surface temperature of 8.70°C, find the temperature (in Celsius) at a depth of 35.0 km (near the base of the crust). Ignore the heat generated by the presence of radioactive elements.

So i understand that the surface is radiating energy at 63.7 mW/m^2
and that k for the surface rocks is 3.96
whilst the surface temperature is 8.70°C


I thought of using P = KA x [(T(h) - T(c)) / L]
L = 35500 M
K = 3.96
P = 0.0637 W
A = 1 m ^ 2
T(h) - T(c) = 8.60 - T(c)

Solving for T(c) = -562.45 <---- I don't think that's correct at all

would any of you offer me in assistance on how to approach this question?
 
Physics news on Phys.org
How is 8.6 T(h)? It's hotter deeper down.
 
haha silly me,
hence if i use T(c) as 8.6
then i get T(H) as 579.65 which I can understand because it gets quite hot down there, but would that be correct?
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top