Question

The heat generated by radioactivity within the earth is conducted outward through the oceans. Assuming the average temperature gradient within the solid earth beneath the ocean to be ${0.07}^{\circ }{C}^{-1}$ and the average thermal conductivity $0.2cal{m}^{-1}{s}^{-1}{C}^{-1}$, determine the rate of heat transfer per square meter. Radius of the earth = 6400 km. Further, determine the quantity of heat transferred through the earth's surface each day.

• A. $8.2\times {10}^{17}cal$
• B. $6.2\times {10}^{17}cal$
• C. $16.92\times {10}^{17}cal$
• D. $69\times {10}^{17}cal$

Answer 1

The correct option is B $6.2\times {10}^{17}cal$

From the previous problem
$Q=\frac{4\pi \lambda (T_1-T_2)r_1{r}_2}{(r_1-r_2)}\beta \left(temperaturegradient\right)=\frac{T_1-T_2}{r_1-r_2}\therefore Q=4\pi \lambda \beta r_1{r}_{2}Here{r}_1=r_{2}r\left(radiusoftheearth\right)\therefore Q=4\pi \lambda \beta r^{2}Rateofheattransferthroughunitarea=\frac{4\pi \lambda \beta r^2}{4\pi r^2}=\lambda \beta =0.2\times 0.07=14\times {10}^{-3}calHeattransferperday=14\times {10}^{-3}\times 4\pi (6.4\times {10}^6)\times 86400=6.2\times {10}^{17}cal$