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Q. 3.13

Expert-verifiedFound in: Page 97

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

When the sun is high in the sky, it delivers approximately 1000 watts of power to each square meter of earth's surface. The temperature of the surface of the sun is about $6000\mathrm{K}$, while that of the earth is about $300\mathrm{K}$.

(a) Estimate the entropy created in one year by the flow of solar heat onto a square meter of the earth.

(b) Suppose you plant grass on this square meter of earth. Some people might argue that the growth of the grass (or of any other living thing) violates the second law of thermodynamics, because disorderly nutrients are converted into an orderly life form. How would you respond?

(a) The change in entropy can be estimated to be $4.97\times {10}^{7}J{K}^{-1}$.

(b) The growth of grass doesn't violate the second law of thermodynamics as the entropy produced to transmit the energy from the sun to earth (disordered) would be much higher when compared to the entropy produced by the biochemical reactions (ordered) in the grass.

Power delivered per square meter$={P}_{s}=1000W$

Surface temperature of sun $={T}_{sun}=6000K$

Surface temperature of earth ${T}_{earth}=300K$

Time period (day time) role="math" localid="1647243842287" $=t=1year=12\times 3600\times 365s$

Amount of energy received by the earth per square meter can be calculated as:

$Q={P}_{s}\times t$

By substituting the values, we get,

$Q=1000\times (12\times 3600\times 365)\phantom{\rule{0ex}{0ex}}Q=1.57\times {10}^{10}\mathrm{J}$

Now the changes in entropies can be calculated as:

For earth:

role="math" localid="1647244335644" $\Delta {S}_{earth}=\frac{Q}{{T}_{earth}}$

By substituting the values, we get,

$\Delta {S}_{\text{earth}}=\frac{1.57\times {10}^{10}}{300}\phantom{\rule{0ex}{0ex}}\Delta {S}_{\text{earth}}=5.23\times {10}^{7}J{K}^{-1}$

For sun:

role="math" localid="1647244543097" $\Delta {S}_{sun}=-\frac{Q}{{T}_{sun}}$

By substituting the values, we get,

$\Delta {S}_{\text{sun}}=-\frac{1.57\times {10}^{10}}{6000}\phantom{\rule{0ex}{0ex}}\Delta {S}_{\text{sun}}=-2.62\times {10}^{6}J{K}^{-1}$

We know that the net entropy change can be calculated as:

$\Delta {S}_{\text{total}}=\Delta {S}_{\text{sun}}+\Delta {S}_{\text{earth}}$

By substituting the calculated values, we get,

$\Delta {S}_{\text{total}}=5.23\times {10}^{7}-2.62\times {10}^{6}\phantom{\rule{0ex}{0ex}}\Delta {S}_{\text{total}}=4.97\times {10}^{7}{\mathrm{JK}}^{-1}$

Hence, the total entropy change can be calculated as $4.97\times {10}^{7}J{K}^{-1}$.

Entropy is a measure of a system's randomness or disorder in general. The total entropy can't be reduced, but it can stay the same if the processes are reversible, according to the law of thermodynamics.

The second rule of thermodynamics is not violated by growing grass. This is due to its ability to harness solar energy and transform disordered molecules in the soil into an organized life. However, in the event of a biological reaction, entropy would be substantially lower, and hence far lower than the entropy caused by transporting energy from the Sun to the Earth.

The entropy produced by transmitting energy from the sun to earth (disordered) would be far more than the entropy produced by biological reactions (ordered) in the grass, hence the development of grass does not break the second law of thermodynamics.

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