Question

Find the change in entropy of the H2O molecules when 1.38 kilograms of ice melts into water at 273 K. J/K (b) Find the change in entropy of the H2O molecules when 1.38 kilograms of water changes into steam at 373 K. J/K (c) On the basis of the answers to parts (a) and (b), discuss which change creates more disorder in the collection of H2O molecules.

Answer 1

Answer:

(a) $\Delta s_i=1718.68\,J.K^{-1}$

(b) $\Delta s_v=8509.38\,J.K^{-1}$

(c) Vaporization creates more disorder in the water molecules.

Explanation:

Given that:

mass of water, $m=1.38kg$

transition temperature of ice into water, $T_i=273K$

transition temperature of water into steam, $T_v=373K$

We know:

Latent heat of fusion of ice, $L_i=3.4\times 10^5 J.kg^{-1}$

Latent heat of vapourization of water, $L_v=2.3\times 10^6 J.kg^{-1}$

Heat change during melting of ice:

$dQ_i=m.L_i$

$dQ_i=1.38\times 3.4\times 10^5$

$dQ_i=4.692\times 10^5\,J$

Heat change during vaporization:

$dQ_v=m.L_v$

$dQ_v=1.38\times 2.3\times 10^6$

$dQ_v=3.174\times 10^6\,J$

we have the equation of entropy change Δs:

$\Delta s=\frac{dQ}{T}$

where:

T= temperature in kelvin

dQ= change in heat energy

(a)

∴The change in entropy of the water molecules when 1.38 kilograms of ice melts into water at 273 K

$\Delta s_i=\frac{dQ_i}{T_i}$.....................()

$\Delta s_i=\frac{4.692\times 10^5}{273}$

$\Delta s_i=1718.68\,J.K^{-1}$

(b)

∴The change in entropy of the water molecules when 1.38 kilograms of water vaporizes at 373 K

$\Delta s_v=\frac{dQ_v}{T_v}$.....................()

$\Delta s_v=\frac{3.174\times 10^6}{373}$

$\Delta s_v=8509.38\,J.K^{-1}$

(c)

Vapourization of water creates more randomness and disorderliness in the molecules of the water because change in entropy is greater for vapourization and entropy change is the denotes randomness.