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 $745.65 J K^{- 1}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

Amount of hot water $= V_{1} = 50 L = 50000 g$

Temperature of hot water $= T_{1} = 55 ° C = 328 K$

Amount of cold water $= V_{2} = 25 L = 25000 g$

Temperature of cold water $= T_{2} = 10 ° C = 283 K$

Step 2: Calculation

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

$T_{f} = \frac{(T_{1} \times V_{1}) + (T_{2} \times V_{2})}{V_{1} + V_{2}}$

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

$T_{f} = \frac{(328 \times 50) + (283 \times 25)}{50 + 25} T_{f} = 313 K$

The change in entropy is given as:

role="math" localid="1647236968068" $Δ S = C_{V} \int_{T_{i}}^{T_{f}} \frac{1}{T} d T . . . . . . . . . . (1)$

Where,

$C_{V}$ is the heat capacity at constant volume and is given as: $C_{V} = m c$

Where,

$m$ = mass

$c$ = specific heat

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

$Δ S = m c [\ln \frac{T_{f}}{T_{i}}] . . . . . . . . . . (2)$

Now,

for hot water:

$T_{i} = 328 K m = 50000 g$

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

role="math" localid="1647238076045" $Δ S_{h o t} = 50000 \times 4.18 \times [\ln \frac{313}{328}] Δ S_{h o t} = - 9783.38 J K^{- 1}$

for cold water:

$T_{i} = 283 K m = 25000 g$

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

$Δ S_{c o l d} = 25000 \times 4.18 \times [\ln \frac{313}{283}] Δ S_{c o l d} = 10529.03 J K^{- 1}$

Thus, the net entropy change can be given as:

$Δ S = Δ S_{h o t} + Δ S_{c o l d} Δ S = - 9783.38 + 10529.03 Δ S = 745.65 J K^{- 1}$

Step 3: Final answer

Hence, the required change in entropy can be calculated as $745.65 J K^{- 1}$ .

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