Entropy Associated with Disorder

Find the increase in entropy of 1.00 kg of ice that is originally at $0 °C$ and melts to form water at $0 °C$ .

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The change in entropy can be calculated from the definition of $Δ S$ once we find the energy, Q, needed to melt the ice.
Solution
The change in entropy is defined as

$Δ S = \frac{Q}{T} .$
12.15

Here, Q is the heat necessary to melt 1.00 kg of ice and is given by

$Q = m L_{f},$
12.16

where m is the mass and $L_{f}$ is the latent heat of fusion. $L_{f} = 334 kJ/kg$ for water, so

$Q = (1 .00 kg) (334 kJ/kg) = 3.34 \times 10^{5} J.$
12.17

Because Q is the amount of energy heat adds to the ice, its value is positive, and T is the melting temperature of ice, $T = 273 K$ So the change in entropy is

$Δ S = \frac{Q}{T} = \frac{3.34 \times 10^{5} J}{273 K} = 1.22 \times 10^{3} J/K.$
12.18

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