Question

Short answer

The heat transfer from the interior to the surface is (a)$4.9\times {10}^{16}\mathrm{J}$, and the comparison result is (b)$Q^{\prime }=1.26\times {10}^{4}Q$.

Step-by-step solution

Given data

The temperature is$\mathrm{\Delta }T=1^{\circ }\mathrm{C}$.

The depth is$d=30\mathrm{m}$.

The thermal conductivity is$k=0.80\mathrm{J}/\mathrm{s}{\cdot }^{\circ }\mathrm{C}\cdot \mathrm{m}$.

The time is$t=1\mathrm{h}$.

The heat on the Earth’s surface is$I=1000\mathrm{W}/{\mathrm{m}}^2$.

Understanding heat transfer

Consider that all of the heat is transferred to the surface of the Earth by passing 30 m of depth. Use the standard value of the radius of Earth while performing the calculation.

Calculation of the heat transferred

The relation to calculate the heat transfer is given by:

$\begin{array}{r}l\frac{Q}{t}=kA\frac{\mathrm{\Delta }T}{d}\\ Q=\left(k\left(4\pi R^2\right)\frac{\mathrm{\Delta }T}{d}\right)t\end{array}$

Here,$A$is the area of Earth and Ris the radius of Earth.

On plugging the values in the above relation, you get:

$\begin{array}{r}lQ=\left(\left(0.80\mathrm{J}/\mathrm{s}{\cdot }^{\circ }\mathrm{C}\cdot \mathrm{m}\right)\left(4\pi {\left(6.38\times {10}^6\mathrm{m}\right)}^2\right)\frac{1^{\circ }\mathrm{C}}{30\mathrm{m}}\right)\left(1\mathrm{h}\times \frac{3600\mathrm{s}}{1\mathrm{h}}\right)\\ Q=4.9\times {10}^{16}\mathrm{J}\end{array}$

Thus, $Q=4.9\times {10}^{16}\mathrm{J}$ is the required heat transfer.

Calculation of the heat coming from the Sun and its comparison with the heat transferred from Earth’s surface  

The relation to calculate the energy incident on the Earth is given by:

$\begin{array}{r}l\frac{Q^{\prime }}{t}=\sigma A^{\prime }\\ Q^{\prime }=\sigma A^{\prime }t\end{array}$

Here,$A^{\prime }$is the area of Sun and$\sigma$is the solar constant.

On plugging the values in the above relation, you get:

$\begin{array}{r}l{Q}^{\prime }=\left(1350\mathrm{W}/{\mathrm{m}}^2\right)\left(\pi {\left(6.38\times {10}^6\mathrm{m}\right)}^2\right)\left(1\mathrm{h}\times \frac{3600\mathrm{s}}{1\mathrm{h}}\right)\\ Q^{\prime }=6.21\times {10}^{20}\mathrm{J}\end{array}$

The comparison between the two heat transfers is:

$\begin{array}{r}l\frac{Q^{\prime }}{Q}=\frac{6.21\times {10}^{20}\mathrm{J}}{4.9\times {10}^{16}\mathrm{J}}\\ Q^{\prime }=1.26\times {10}^{4}Q\end{array}$

Thus, $Q^{\prime }=1.26\times {10}^{4}Q$ is the comparison between the heat transfer from the Sun and the interior of Earth.