Question 16.4: Entropy change in melting ice Now let’s apply our concept of...
Entropy change in melting ice
Now let’s apply our concept of entropy to a well-defined transformation from an ordered state to a disordered one. In particular, we will consider the melting of ice. Ice has an ordered crystal structure that has a relatively low entropy. Melting the ice into water destroys the crystalline arrangement of the water molecules, thereby increasing the entropy of the system. We will use Equation 16.10 to compute the change in entropy of 1.00 kg of ice at 0°C when it is melted and converted to water at 0°C.
SET UP The temperature T is constant at 273 K, and the heat of fusion Q needed to melt the ice is \mathrm{Q = mL_f = (1.00 kg) (334 \times 10^3 J/kg) = 334 \times 10^3 J.}
SOLVE From Equation 16.10, the increase in entropy of the system is
\mathrm{\Delta S=\frac{Q}{T}=\frac{3.34 \times 10^5 J}{273 K}=1220 J/K. }
REFLECT The entropy of the system increases as heat is added to it.
The disorder increases because the water molecules in the liquid state have more freedom to move than when they were locked in a crystal structure. In any isothermal reversible process, the change of entropy equals the heat transferred divided by the absolute temperature. When we refreeze the water, its entropy change is ΔS = -1220 J/K.
Practice Problem: When 50 g of liquid water at 0°C freezes into a 50 g ice cube at 0°C, what is the change in entropy of the material? Answer: -61 J/K.