An ice cube (mass 30 g) at 0°C is left sitting on the

An Introduction to Thermal Physics | 1st Edition | ISBN: 9780201380279 | Authors: Daniel V. Schroeder

Problem 10P Chapter 3

An Introduction to Thermal Physics | 1st Edition

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An Introduction to Thermal Physics | 1st Edition | ISBN: 9780201380279 | Authors: Daniel V. Schroeder

An Introduction to Thermal Physics | 1st Edition

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Problem 10P

An ice cube (mass 30 g) at 0°C is left sitting on the kitchen table, where it gradually melts. The temperature in the kitchen is 25°C.

(a) Calculate the change in the entropy of the ice cube as it melts into water at 0°C. (Don’t worry about the fact that the volume changes somewhat.)

(b) Calculate the change in the entropy of the water (from the melted ice) as its temperature rises from 0°C to 25°C.

(c) Calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water.

(d) Calculate the net change in the entropy of the universe during this process. Is the net change positive, negative, or zero? Is this what you would expect?

Step-by-Step Solution:

Part (a)

        To calculate the change in the entropy of the ice cube  as it melts into water at 0°C

Step 1:

        Mass of the ice cube m = 30 g or 0.03 kg

        Temperature of ice cube TC = 0°C or 273 K

        Temperature of the kitchen TH = 25°C or 298 K

Step 2:

The entropy of the ice cube (constant temperature)

=

Where Q is the energy needed to melt the ice

Q1 = m x Lf

Lf is the latent heat of fusion of ice

Lf  = 334 KJ/K

Step 3:

        =

        =

        = 36.7 J/K

        The change in the entropy of the ice cube is 36.7 J/K

Part (b)

Step 1:

        To calculate the change in the entropy of the water (from the melted ice) as its temperature rises from 0°C to 25°C.

        Mass of the ice cube m = 30 g or 0.03 kg

        Temperature of ice cube TC = 0°C or 273 K

        Temperature of the kitchen TH = 25°C or 298 K

Step 2:

        Here the temperature changes and the change in entropy is

        =

        Where c is the specific heat of water

        c = 4190 J/K

        =

        =

= 11.01 J/K

The change in the entropy of the water is 11.01 J/K

Part (c)

Step 1:

To calculate the change in the entropy of the kitchen as it gives up heat to the melting ice/water

Mass of the ice cube m = 30 g or 0.03 kg

        Temperature of ice cube TC = 0°C or 273 K

Temperature of the kitchen TH = 25°C or 298 K

Step 2:

        The amount of heat transferred from the kitchen to the ice Q = Q1 + Q2

        Q1 - heat required to melt the ice

        Q1 - heat required to increase the temperature from  0°C to 25°C

        

Q1 =  m x Lf =

        Q1 = 10020 J

        

        Q2 =  mc(TH - TC)

        Q2 = 0.03*4190*(298-273)

        Q2 = 3142.5 J

        Q = 10020 J + 3142.5 J = 13162.5 J

        The amount of heat transferred from the kitchen to the ice Q = 13162.5 J

Step 3:

        The change in entropy of the room happens at constant room temperature. The change in the entropy of the kitchen

        = =

        = -44.17 J/K

        The change in the entropy of the kitchen is -44.17 J/K

Part (d)

Step 1:

        To find the net change in the entropy of the universe during this process

        = + +

        = 36.7 J/K + 11.01 J/K + (-44.17 J/K)

        = + 3.54 J/K

        The net change in the entropy of the universe during this process is + 3.54 J/K

Step 2 of 2

Chapter 3, Problem 10P is Solved
Textbook: An Introduction to Thermal Physics
Edition: 1st
Author: Daniel V. Schroeder
ISBN: 9780201380279

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