# Thermal conductivity of earth

1. Oct 16, 2014

### toothpaste666

1. The problem statement, all variables and given/known data
The temperature within the Earth's crust increases about 1.0 C∘ for each 30 m of depth. The thermal conductivity of the crust is 0.80 W/C∘⋅m.
A)Determine the heat transferred from the interior to the surface for the entire Earth in 9.0h .
B)Compare this heat to the amount of energy incident on the Earth in 9.0h due to radiation from the Sun.

2. Relevant equations
dQ/dt = kA(dT/l)

3. The attempt at a solution
I have solved the first part.
surface area of earth = 4pir^2 = 4pi(6.378x10^6 m)^2 = 5.1x10^14 m^2
dQ/dt = kA(dT/l)
dQ/dt = (.80 J/smC)(5.1x 10^14 m^2)(1 C/ 30m)
dQ/dt = 1.36x10^13 J/s

9h is 9h(3600 s/h) = 32400 s

32400s(1.36x10^13 J/s) = 4.4x10^17 J

The second part I am having a hard time understanding. I'm not really sure I even know what they are asking

2. Oct 17, 2014

### BvU

Well, I take it they want you to estimate solar irradiance, right ?
Don't worry about more than one or two decimals of accuracy...
And pay attention to the fact that values such as in the table here are for perpendicular incidence.

3. Mar 10, 2017

### conner

part b of the problem asks you to compare energy into the surface from the core and compare it to the sun,
energy from the sun:
1300 W/m^2 *(4*pi*(6.37*10^6)^2)* .5 (only half the earth is hit per second) = 3.32*10^17 W sun.
3.32*10^17 W * 32400 s = 1.07*10^22 J
now create a ratio:
Jcore/Jsun = (4.4*10^17)/(1.07*10^22)

the earths core only adds .004% of the suns solar intensity to the surface.