Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

(a) On a winter day, a certain house loses 5.00×10⁸ J of heat to the outside (about 500,000 Btu). What is the total change in entropy due to this heat transfer alone, assuming an average indoor temperature of 21.0 ºC and an average outdoor temperature of 5.00 ºC ? (b) This large change in entropy implies a large amount of energy has become unavailable to do work. Where do we find more energy when such energy is lost to us?

The total entropy change is 98000 J/K. More energy can be found by finding things that would keep the house and us warm.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Introduction

The entropy represents the randomness of a thermodynamic system. It is the measure of disorder in a system.

Step 2: Given parameters and formula for change in entropy

Heat transferred is 5×10⁸J

Temperature indoor T_h= 21 ̊ C = 294 K

Temperature outdoor T_c₌5 ̊ C = 278K

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Here,

\[\Delta S\]- change in entropy.

Q - heat transferred.

T- temperature.

Step 3: Calculate the entropy change

Change in indoor entropy

∆ S_h= - Q/T_h

Change in outdoor entropy

∆S_c= Q/T_c

Total change in entropy

\begin{aligned} \Delta {S_{total}} = \Delta {S_h} + \Delta {S_c}\\ = - Q/{T_h} + Q/{T_c}\\= Q\left( {\frac{{ - 1}}{{294\;{\rm{K}}}} + \frac{1}{{278\;{\rm{K}}}}} \right)\\= 98000\;{\rm{J/K}}\end{aligned}

Therefore, the total change in entropy is 98000 J/K. More energy can be found by finding things that would keep the house and us warm like a heater, fire, woolen clothes, etc. We just need things that would provide us energy or keep energy trapped to perform work

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Step 1: Introduction

Step 2: Given parameters and formula for change in entropy

Step 3: Calculate the entropy change

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College Physics (Urone)

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Step 1: Introduction

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Step 3: Calculate the entropy change

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