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Q. 3.11

An Introduction to Thermal Physics
Found in: Page 97
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

In order to take a nice warm bath, you mix 50 liters of hot water at 55°C with 25 liters of cold water at 10°C. How much new entropy have you created by mixing the water?

By mixing the water, the change in entropy is 745.65 JK-1.

See the step by step solution

Step by Step Solution

Step 1: Given

Amount of hot water =V1=50 L=50000 g

Temperature of hot water =T1=55°C=328 K

Amount of cold water =V2=25 L=25000 g

Temperature of cold water =T2=10°C=283 K

Step 2: Calculation

Since the hot and cold waters are mixed together, the final temperature of the water can be given as:


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

Tf=(328×50)+(283×25)50+25Tf=313 K

The change in entropy is given as:

role="math" localid="1647236968068" ΔS=CVTiTf1TdT ..........(1)


CV is the heat capacity at constant volume and is given as:CV=mc


m = mass

c = specific heat

Hence, equation (1) can be written in a simplified way as:

ΔS=mclnTfTi ..........(2)


for hot water:

Ti=328 Km=50000 g

By substituting the values in equation (2), we get,

role="math" localid="1647238076045" ΔShot=50000×4.18×ln313328ΔShot=-9783.38 JK-1

for cold water:

Ti=283 Km=25000 g

By substituting the values in equation (2), we get,

ΔScold=25000×4.18×ln313283ΔScold=10529.03 JK-1

Thus, the net entropy change can be given as:

ΔS=ΔShot+ΔScoldΔS=-9783.38+10529.03ΔS=745.65 JK-1

Step 3: Final answer

Hence, the required change in entropy can be calculated as 745.65 JK-1.

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