Respuesta :
Answer:
(a) [tex]\Delta s_i=1718.68\,J.K^{-1}[/tex]
(b) [tex]\Delta s_v=8509.38\,J.K^{-1}[/tex]
(c) Vaporization creates more disorder in the water molecules.
Explanation:
Given that:
mass of water, [tex]m=1.38kg[/tex]
transition temperature of ice into water, [tex]T_i=273K[/tex]
transition temperature of water into steam, [tex]T_v=373K[/tex]
We know:
Latent heat of fusion of ice, [tex]L_i=3.4\times 10^5 J.kg^{-1}[/tex]
Latent heat of vapourization of water, [tex]L_v=2.3\times 10^6 J.kg^{-1}[/tex]
Heat change during melting of ice:
[tex]dQ_i=m.L_i[/tex]
[tex]dQ_i=1.38\times 3.4\times 10^5[/tex]
[tex]dQ_i=4.692\times 10^5\,J[/tex]
Heat change during vaporization:
[tex]dQ_v=m.L_v[/tex]
[tex]dQ_v=1.38\times 2.3\times 10^6[/tex]
[tex]dQ_v=3.174\times 10^6\,J[/tex]
we have the equation of entropy change Δs:
[tex]\Delta s=\frac{dQ}{T}[/tex]
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
[tex]\Delta s_i=\frac{dQ_i}{T_i}[/tex].....................()
[tex]\Delta s_i=\frac{4.692\times 10^5}{273}[/tex]
[tex]\Delta s_i=1718.68\,J.K^{-1}[/tex]
(b)
∴The change in entropy of the water molecules when 1.38 kilograms of water vaporizes at 373 K
[tex]\Delta s_v=\frac{dQ_v}{T_v}[/tex].....................()
[tex]\Delta s_v=\frac{3.174\times 10^6}{373}[/tex]
[tex]\Delta s_v=8509.38\,J.K^{-1}[/tex]
(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.
