Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

An 8.0 g ice cube at $- 10 ° C$ is put into a Thermos flask containing $100 {cm}^{3}$ of water at $20 ° C$ . 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 $0.64 J / ° C$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

Mass of ice is $m_{i c e} = 8 g m (\frac{10^{- 3} kg}{1 gm}) = 0.008 kg$

Initial Temperature of ice $T_{initial}^{ice} = - 10 ° C = 263.15 kg$

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

Mass of water $m_{water} = 100 gm (\frac{10^{- 3} kg}{1 gm}) = 0.1 kg$

Initial Temperature of water $T_{initial}^{water} = 20 ° C = 293.15 K$

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,

$∆ S_{temp change} = mc In (\frac{T_{final}}{T_{initial}})$ (1)

$∆ S = \frac{L_{F} m}{T}$

The heat transferred by the body,

$Q = cm ∆ T$ (2)

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

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

For $m_{water} = 0.1 kg$ of water, the specific heat is given by

$c_{water} = 4.190 J / g . ° C = 4190 J / kg . ° C$

When the equilibrium is reached, then $\sum_{} Q = 0$

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

$Q_{warm water cools} + Q_{ice warms to 0 ° C} + Q_{ice melts} + Q_{melted ice warms = 0}$

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

$c_{w a t e r} m_{w a t e r} (T_{f} - 20 ° C) + c_{i c e} m_{i c e} (0 ° C - (- 10 ° C)) + L_{F} m_{i c e} + c_{w a t e r} m_{i c e} (T_{f} - 0 ° C) = 0$

$L_{F}$ is heat fusion of water, and its value is given by $L_{F} = 333 \times 10^{3} \frac{J}{k g}$

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

$[(4190 J / kg . ° C \times 0.1 kg) T_{f} - (4190 J / kg . ° C \times 0.1 kg \times 20 ° C)] + [2220 J / kg . ° C \times 0.008 kg \times 10 ° C] + (333 \times 10^{3} J / kg \times 0.008 kg) + (4190 J / kg . ° C \times 0.008 kg \times T_{f}) = 0$

$(419 J / ° C) T_{f} - 8380 J + 177.6 J + 2841.6 J + (33.52 J / ° C) T_{f} = 0 (452.52 J / ° C) T_{f} = 5538.4 J T_{f} = 12.24 ° C T_{f} = 285.39 K$

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

$∆ S_{temp change} = mcIn (\frac{T_{2}}{T_{1}}) ∆ S_{melt} = \frac{L_{F} m}{T_{0}}$

For the phase change experienced by the ice $(w i t h T_{0} = 273.15 K)$ the total entropy change is given as:

$∆ S_{system} = m_{water} c_{water} In (\frac{285.39 K}{293.15 K}) + m_{ice} c_{ice} In (\frac{273.15 K}{263.15 K}) + m_{ice} c_{water} In (\frac{285.39 K}{273.15 K}) + (\frac{L_{F} m_{ice}}{273.15 K}) = (0.1 kg) (4190 J / kg . ° C) In (\frac{285.39 K}{293.15 K}) + (0.008 kg) (2220 J / kg . ° C) In (\frac{285.39 K}{273.15 K}) + (\frac{2664 J}{273.15 K})$

$∆ S_{s y s t e m} = (- 11.24 + 0.66 + 1.47 + 9.75) J / ° C = 0.64 J / ° C$

Hence, the value of the entropy change of the system is $0.64 J / ° C$

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Fundamentals Of Physics

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Short Answer

An 8.0 g ice cube at $- 10 ° C$ is put into a Thermos flask containing $100 {cm}^{3}$ of water at $20 ° C$ . 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

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of entropy change

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

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An 8.0 g ice cube at -10°C is put into a Thermos flask containing 100 cm3of water at 20°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

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of entropy change

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

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An 8.0 g ice cube at $- 10 ° C$ is put into a Thermos flask containing $100 {cm}^{3}$ of water at $20 ° C$ . 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