Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

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 $8.0 \times 10^{4} J / K$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

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 $= P = 4 k W = 4 \times 10^{3} J / s$

Temperature inside $= T_{i n} = 293 K$

Temperature outside $= T_{o u t} = 275 K$

Step 2: Calculation

Total heat loss in a day can be calculated as:

$Q = P t$

Where,

$P$ = Power

$t$ = time

Hence,

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

Now,

Entropy gained by outdoors can be given as:

$Δ S_{out} = \frac{Q}{T_{out}}$

By substituting the values, we get,

$Δ S_{out} = \frac{3.46 \times 10^{8}}{275} Δ S_{out} = 1.26 \times 10^{6} J / K$

And,

Entropy gained by indoors can be given as:

$Δ S_{i n} = - \frac{Q}{T_{i n}}$

By substituting the values, we get,

$Δ S_{in} = - \frac{3.46 \times 10^{8}}{293} Δ S_{i n} = - 1.18 \times 10^{6} J / K$

We know that the net entropy change is given as:

$Δ S_{n e t} = Δ S_{o u t} + Δ S_{in}$

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

$Δ S_{net} = (1.26 \times 10^{6}) - (1.18 \times 10^{6}) Δ S_{net} = 8.0 \times 10^{4} J / K$

Step 3: Final answer

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

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