Question

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 9.0 h . (b)Compare this heat to the amount of energy incident on the Earth in 9.0 h due to radiation from the Sun. Express your answer using two significant figures.

Answer 1

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

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.