Q. The temperature gradient in the earth’s crust is 32$℃$ k$m^{- 1}$ and the mean conductivity of earth is 0.008 cal$s^{- 1}cm^{- 1}℃^{- 1}$. Considering earth to be a sphere of radius 6000 km loss of heat by earth everyday is about
A) $10^{30}$ cal
B) $10^{40}$ cal
C) $10^{20}$ cal
D) $10^{18}$ cal
Explanation
(d) $\mathrm{\Delta}Q = KA\left( \frac{\mathrm{\Delta}T}{\mathrm{\Delta}x} \right)\mathrm{\Delta}t,\ \ where\ A = 4\ \pi\ r^{2}$
= 0.008$\times 4\ \times \frac{22}{7}\left( 6 \times 10^{8} \right)^{2} \times \left( \frac{32}{10^{5}} \right) \times 86400$
= $10^{18}$ cal
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.