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Q79E

Expert-verified
Found in: Page 67

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

At Earth's location, the intensity of sunlight is 1.5 kW / ${{\mathbf{m}}}^{{\mathbf{2}}}$ . If no energy escaped Earth, by how much would Earth's mass increase in 1 day?

The increase in Earth’s mass for no escape of Sun’s energy from Earth is $7.34×{10}^{5}\mathrm{kg}$ .

See the step by step solution

Step 1: Identification of given data

The intensity of sunlight on Earth is $\mathrm{I}=1.5\mathrm{kW}/{\mathrm{m}}^{2}$ .

The number of seconds in a day on Earth is $\mathrm{t}=1\mathrm{day}=86400\mathrm{s}$.

The energy received by Earth per unit surface area from the sun is called the solar constant.

Step 2: Determination of energy received by Earth in one day

The energy received by Earth in time t is given as:

$E=I.A.t$

Here, A is the surface area of Earth and its value is $5.10×{10}^{14}{\mathrm{m}}^{2}$.

For the given values, equation becomes-

$E=\left(1.5\mathrm{kW}/{\mathrm{m}}^{2}\right)\left(\frac{{10}^{3}\mathrm{J}/\mathrm{s}}{1\mathrm{kW}}\right)\left(5.10×{10}^{14}\right)\left(86400\mathrm{s}\right)\phantom{\rule{0ex}{0ex}}E=6.61×{10}^{22}\mathrm{J}$

Step 3: Determination of increase in Earth’s mass

The increase in Earth’s mass is given as:

$E=m{c}^{2}$

Here, c is the speed of light in vacuum and its value is $3×{10}^{8}\mathrm{m}/\mathrm{s}$

$6.61×{10}^{22}\mathrm{J}=\mathrm{m}{\left(3×{10}^{8}\mathrm{m}/\mathrm{s}\right)}^{2}\phantom{\rule{0ex}{0ex}}m=7.34×{10}^{5}\mathrm{kg}$

Therefore, the increase in Earth’s mass for no escape of energy from Earth is $7.34×{10}^{5}\mathrm{kg}$ .