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

## Problem 10P Chapter 3

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

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