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Q. 3.12

Expert-verifiedFound in: Page 97

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

Estimate the change in the entropy of the universe due to heat escaping from your home on a cold winter day.

The entropy change on a cold winter day can be estimated to be $8.0\times {10}^{4}\mathrm{J}/\mathrm{K}$.

It is given to estimate the change in the entropy of the universe due to heat escaping from your home on a cold winter day.

Hence,

Let's assume:

Power consumed from an average house on a winter day $=P=4kW=4\times {10}^{3}J/s$

Temperature inside $={T}_{in}=293K$

Temperature outside $={T}_{out}=275K$

Total heat loss in a day can be calculated as:

$Q=Pt$

Where,

$P$ = Power

$t$ = time

Hence,

$Q=4\times {10}^{3}\times 24\times 60\times 60\phantom{\rule{0ex}{0ex}}Q=3.46\times {10}^{8}J$

Now,

Entropy gained by outdoors can be given as:

$\Delta {S}_{\text{out}}=\frac{Q}{{T}_{\text{out}}}$

By substituting the values, we get,

$\Delta {S}_{\text{out}}=\frac{3.46\times {10}^{8}}{275}\phantom{\rule{0ex}{0ex}}\Delta {S}_{\text{out}}=1.26\times {10}^{6}\mathrm{J}/\mathrm{K}$

And,

Entropy gained by indoors can be given as:

$\Delta {S}_{in}=-\frac{Q}{{T}_{in}}$

By substituting the values, we get,

$\Delta {S}_{\text{in}}=-\frac{3.46\times {10}^{8}}{293}\phantom{\rule{0ex}{0ex}}\Delta {S}_{in}=-1.18\times {10}^{6}\mathrm{J}/\mathrm{K}$

We know that the net entropy change is given as:

$\Delta {S}_{net}=\Delta {S}_{out}+\Delta {S}_{\text{in}}$

By substituting the calculated values in the above equation, we get,

$\Delta {S}_{\text{net}}=\left(1.26\times {10}^{6}\right)-\left(1.18\times {10}^{6}\right)\phantom{\rule{0ex}{0ex}}\Delta {S}_{\text{net}}=8.0\times {10}^{4}J/K$

Hence, the required entropy change can be calculated as $8.0\times {10}^{4}J/K$.

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