Calculate the daily loss of energy by the earth, if the temperature gradient in the earth's crust is `32^(@)C` per km and mean conductivity of the rock is `0.008` of CGS unit. (Given radius of earth `=6400km`)

Options

A. `10^(40)cal`
B. `10^(30)cal`
C. `10^(18)cal`
D. `10^(10)cal`

Explanation

(c) Temperature gradient `(d theta)/(dx)=32/1000.^(@)C//m=32/(10^(5)).^(@)C//m`
Loss of energy by earth is given by
`Q=KA(d theta)/(dx)xx86400`
`=0.008xx(4xx22)/7xx(64xx10^(8))^(2)xx32/(10^(5))xx86400`
`=1.1xx10^(18)cal=10^(18)`cal