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Found in: Page 605

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# An 8.0 g ice cube at ${\mathbf{-}}{\mathbf{10}}{\mathbf{°}}{\mathbf{C}}$ is put into a Thermos flask containing ${\mathbf{100}}{\mathbf{}}{{\mathbf{cm}}}^{{\mathbf{3}}}$of water at ${\mathbf{20}}{\mathbf{°}}{\mathbf{C}}$. By how much has the entropy of the cube – water system changed when equilibrium is reached? The specific heat of ice is 2220J/kg.K

Entropy change of the cube-water system when equilibrium is reached is $0.64\mathrm{J}/°\mathrm{C}$.

See the step by step solution

## Step 1: The given data

Mass of ice is${m}_{ice}=8gm\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{gm}}\right)=0.008\mathrm{kg}$

Initial Temperature of ice${\mathrm{T}}_{\mathrm{initial}}^{\mathrm{ice}}=-10°\mathrm{C}=263.15\mathrm{kg}$

Specific heat of ice${\mathrm{c}}_{\mathrm{ice}}=2220\mathrm{J}/\mathrm{lg}.\mathrm{K}=2220\mathrm{J}/\mathrm{kg}.°\mathrm{C}$

Mass of water${\mathrm{m}}_{\mathrm{water}}=100\mathrm{gm}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{gm}}\right)=0.1\mathrm{kg}$

Initial Temperature of water ${\mathrm{T}}_{\mathrm{initial}}^{\mathrm{water}}=20°\mathrm{C}=293.15\mathrm{K}$

## Step 2: Understanding the concept of entropy change

We use the concept of entropy change of the two blocks coming to thermal equilibrium to calculate entropy. We use the formula of specific heat of the water to calculate the final temperature of the water.

Formulae:

The entropy change of the gas,

$∆{\mathrm{S}}_{\mathrm{temp}\mathrm{change}}=\mathrm{mc}\mathrm{In}\left(\frac{{\mathrm{T}}_{\mathrm{final}}}{{\mathrm{T}}_{\mathrm{initial}}}\right)$ (1)

$∆\mathrm{S}=\frac{{\mathrm{L}}_{\mathrm{F}}\mathrm{m}}{\mathrm{T}}$

The heat transferred by the body,

$\mathrm{Q}=\mathrm{cm}∆\mathrm{T}$ (2)

## Step 3: Calculation of the entropy change of the cube-water system

$0.64\mathrm{J}/°\mathrm{C}$For an equilibrium state, the heat lost by $100{\mathrm{cm}}^{3}\left({\mathrm{m}}_{\mathrm{water}}\right)$ of water is absorbed by ice $\left({\mathrm{m}}_{\mathrm{ice}}\right)$ which melts and reaches temperature ${\mathrm{T}}_{\mathrm{f}}>0$

For ${\mathrm{m}}_{\mathrm{water}}=0.1\mathrm{kg}$ of water, the specific heat is given by

${\mathrm{c}}_{\mathrm{water}}=4.190\mathrm{J}/\mathrm{g}.°\mathrm{C}\phantom{\rule{0ex}{0ex}}=4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}$

When the equilibrium is reached, then

During the equilibrium process, warm water at $20°C$ cools to a final temperature ${\mathrm{T}}_{\mathrm{f}}$, ice at temperature $-10°\mathrm{C}$ warms to $0°C$, and this ice first melts, and then melted ice (water) warms.

${\mathrm{Q}}_{\mathrm{warm}\mathrm{water}\mathrm{cools}}+{\mathrm{Q}}_{\mathrm{ice}\mathrm{warms}\mathrm{to}0°\mathrm{C}}+{\mathrm{Q}}_{\mathrm{ice}\mathrm{melts}}+{\mathrm{Q}}_{\mathrm{melted}\mathrm{ice}\mathrm{warms}=0}$

The specific heat of water is given by the formula of equation (2) as given:

${c}_{water}{m}_{water}\left({T}_{f}-20°C\right)+{c}_{ice}{m}_{ice}\left(0°C-\left(-10°C\right)\right)+{L}_{F}{m}_{ice}+{c}_{water}{m}_{ice}\left({T}_{f}-0°C\right)=0$

${\mathrm{L}}_{\mathrm{F}}$ is heat fusion of water, and its value is given by ${L}_{F}=333×{10}^{3}\frac{J}{kg}$

Substituting all values in the above expression, we get that the final temperature can be obtained as,

$\left[\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.1\mathrm{kg}\right){\mathrm{T}}_{\mathrm{f}}-\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.1\mathrm{kg}×20°\mathrm{C}\right)\right]+\phantom{\rule{0ex}{0ex}}\left[2220\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.008\mathrm{kg}×10°\mathrm{C}\right]+\left(333×{10}^{3}\mathrm{J}/\mathrm{kg}×0.008\mathrm{kg}\right)\phantom{\rule{0ex}{0ex}}+\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}×0.008\mathrm{kg}×{\mathrm{T}}_{\mathrm{f}}\right)=0$

$\left(419\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{f}}-8380\mathrm{J}+177.6\mathrm{J}+2841.6\mathrm{J}+\left(33.52\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{f}}=0\phantom{\rule{0ex}{0ex}}\left(452.52\mathrm{J}/°\mathrm{C}\right){\mathrm{T}}_{\mathrm{f}}=5538.4\mathrm{J}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{\mathrm{f}}=12.24°\mathrm{C}\phantom{\rule{0ex}{0ex}}{\mathrm{T}}_{\mathrm{f}}=285.39\mathrm{K}$

Entropy change of two blocks coming to equilibrium is given by using equation (i) can be given as:

$∆{\mathrm{S}}_{\mathrm{temp}\mathrm{change}}=\mathrm{mcIn}\left(\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}\right)\phantom{\rule{0ex}{0ex}}∆{\mathrm{S}}_{\mathrm{melt}}=\frac{{\mathrm{L}}_{\mathrm{F}}\mathrm{m}}{{\mathrm{T}}_{0}}$

For the phase change experienced by the ice $\left(with{T}_{0}=273.15 K\right)$ the total entropy change is given as:

$∆{\mathrm{S}}_{\mathrm{system}}={\mathrm{m}}_{\mathrm{water}}{\mathrm{c}}_{\mathrm{water}}\mathrm{In}\left(\frac{285.39\mathrm{K}}{293.15\mathrm{K}}\right)+{\mathrm{m}}_{\mathrm{ice}}{\mathrm{c}}_{\mathrm{ice}}\mathrm{In}\left(\frac{273.15\mathrm{K}}{263.15\mathrm{K}}\right)\phantom{\rule{0ex}{0ex}}+{\mathrm{m}}_{\mathrm{ice}}{\mathrm{c}}_{\mathrm{water}}\mathrm{In}\left(\frac{285.39\mathrm{K}}{273.15\mathrm{K}}\right)+\left(\frac{{\mathrm{L}}_{\mathrm{F}}{\mathrm{m}}_{\mathrm{ice}}}{273.15\mathrm{K}}\right)\phantom{\rule{0ex}{0ex}}=\left(0.1\mathrm{kg}\right)\left(4190\mathrm{J}/\mathrm{kg}.°\mathrm{C}\right)\mathrm{In}\left(\frac{285.39\mathrm{K}}{293.15\mathrm{K}}\right)+\phantom{\rule{0ex}{0ex}}\left(0.008\mathrm{kg}\right)\left(2220\mathrm{J}/\mathrm{kg}.°\mathrm{C}\right)\mathrm{In}\left(\frac{285.39\mathrm{K}}{273.15\mathrm{K}}\right)+\left(\frac{2664\mathrm{J}}{273.15\mathrm{K}}\right)$

$∆{S}_{system}=\left(-11.24+0.66+1.47+9.75\right)J/°C\phantom{\rule{0ex}{0ex}}=0.64J/°C$

Hence, the value of the entropy change of the system is $0.64\mathrm{J}/°\mathrm{C}$