The temperature gradient in the earth's crust is 32 ° C km- 1 and the mean conductivity of earth is 0.008 cal s- 1 cm- 1 ° C- 1 . Considering the earth to be a sphere of radius 6000 km , the loss of heat by the earth every day is about

Question

The temperature gradient in the earth's crust is 32 ° C km- 1 and the mean conductivity of earth is 0.008 cal s- 1 cm- 1 ° C- 1 . Considering the earth to be a sphere of radius 6000 km , the loss of heat by the earth every day is about (A) 1030cal (B) 1040cal (C) 1020cal (D) 1018 cal

Tardigrade Admin · Accepted Answer

Δ Q=KA((Δ T/Δ x))Δ t =( text0.008/(10)2)× 4 × (22/7)(6 × (10)8)2× ((32/(10)3))× 86400 ≈1018cal

Q. The temperature gradient in the earth's crust is $32^{\circ} C k m^{- 1}$ and the mean conductivity of earth is $0.008 c a l s^{- 1} c m^{- 1}^{\circ} C^{- 1}$ . Considering the earth to be a sphere of radius $6000 km$ , the loss of heat by the earth every day is about

$1 0^{30} c a l$

$1 0^{40} c a l$

$1 0^{20} c a l$

$1 0^{18} c a l$

$Δ Q = K A (\frac{Δ T}{Δ x}) Δ t$
$= \frac{0.008}{( 10 ) ^{2}} \times 4 \times \frac{22}{7} (6 \times (10)^{8})^{2} \times (\frac{32}{( 10 ) ^{3}}) \times 86400$
$\approx 1 0^{18} c a l$

Q. The temperature gradient in the earth's crust is 32∘Ckm−1 and the mean conductivity of earth is 0.008cals−1cm−1∘C−1 . Considering the earth to be a sphere of radius 6000km , the loss of heat by the earth every day is about

1030cal

1040cal

1020cal

1018cal