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Answers without the blur. Sign up and see all textbooks for free! Q. 3.11

Expert-verified Found in: Page 97 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # In order to take a nice warm bath, you mix 50 liters of hot water at $55°\mathrm{C}$ with 25 liters of cold water at $10°\mathrm{C}$. How much new entropy have you created by mixing the water?

By mixing the water, the change in entropy is $745.65J{K}^{-1}$.

See the step by step solution

## Step 1: Given

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

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

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

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

## Step 2: Calculation

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

${T}_{f}=\frac{\left({T}_{1}×{V}_{1}\right)+\left({T}_{2}×{V}_{2}\right)}{{V}_{1}+{V}_{2}}$

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

${T}_{f}=\frac{\left(328×50\right)+\left(283×25\right)}{50+25}\phantom{\rule{0ex}{0ex}}{T}_{f}=313K$

The change in entropy is given as:

role="math" localid="1647236968068" $\Delta S={C}_{V}{\int }_{{T}_{i}}^{{T}_{f}}\frac{1}{T}dT..........\left(1\right)$

Where,

${C}_{V}$ is the heat capacity at constant volume and is given as:${C}_{V}=mc$

Where,

$m$ = mass

$c$ = specific heat

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

$\Delta S=mc\left[\mathrm{ln}\frac{{T}_{f}}{{T}_{i}}\right]..........\left(2\right)$

Now,

for hot water:

${T}_{i}=328K\phantom{\rule{0ex}{0ex}}m=50000g$

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

role="math" localid="1647238076045" $\Delta {S}_{hot}=50000×4.18×\left[\mathrm{ln}\frac{313}{328}\right]\phantom{\rule{0ex}{0ex}}\Delta {S}_{hot}=-9783.38J{K}^{-1}$

for cold water:

${T}_{i}=283K\phantom{\rule{0ex}{0ex}}m=25000g$

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

$\Delta {S}_{cold}=25000×4.18×\left[\mathrm{ln}\frac{313}{283}\right]\phantom{\rule{0ex}{0ex}}\Delta {S}_{cold}=10529.03J{K}^{-1}$

Thus, the net entropy change can be given as:

$\Delta S=\Delta {S}_{hot}+\Delta {S}_{cold}\phantom{\rule{0ex}{0ex}}\Delta S=-9783.38+10529.03\phantom{\rule{0ex}{0ex}}\Delta S=745.65J{K}^{-1}$

Hence, the required change in entropy can be calculated as $745.65J{K}^{-1}$. ### Want to see more solutions like these? 