Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A 364 g block is put in contact with a thermal reservoir. The block is initially at a lower temperature than the reservoir. Assume that the consequent transfer of energy as heat from the reservoir to the block is reversible. Figure gives the change in entropy $∆ S$ of the block until thermal equilibrium is reached. The scale of the horizontal axis is set by $T_{a} = 280 K and T_{b} = 380 K$ . What is the specific heat of the block?

Specific heat of the block is 450 J/kg K .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

a) Mass of the block, m = 364 g or 0.364 kg

b) The graph of entropy change vs. temperature is given.

c) Temperature of horizontal axis, $T_{a} = 280 K and T_{b} = 380 K$

Step 2: Understanding the concept of entropy change

Entropy change is a phenomenon that quantifies how disorder or randomness has changed in a thermodynamic system. We can write the formula for specific heat by rearranging the formula for entropy change. Then inserting the values obtained from the given graph, we can find the specific heat of the block.

Formula:

The entropy change of the gas, $∆ S = m c \ln (\frac{T_{f}}{T_{i}})$ …(i)

Step 3: Calculation for the specific heat of the block

Using equation (i) and the given values, the specific heat of the block is given as:

(From the graph, we can infer that

$T_{f} = 380 K, T_{i} = 280 K and ∆ S = J / K .$ )

$c = \frac{∆ S}{m l n (\frac{T_{f}}{T_{i}})} = \frac{50 J / K}{(0.364 kg) I n (\frac{380 K}{280 K})} = 449.8 J / kgK ~ 450 J / kgK$

Hence, the value of the specific heat of the block is $450 J / kgK$

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of entropy change

Step 3: Calculation for the specific heat of the block

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