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?

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