Answer:
Total change in entropy of the system is 85.84 J/K
Explanation:
Δ[tex]s_{t}[/tex] = Δ[tex]s_{1}[/tex]+Δ[tex]s_{2}[/tex]+Δ[tex]s_{3}[/tex]+Δ[tex]s_{4}[/tex]
Δ[tex]s_{t}[/tex] = total entropy of the system in J/K
Δ[tex]s_{1}[/tex] = entropy due to warming ice from - 10°C (263 K) to 0°C (273 K) in J/K
Δ[tex]s_{2}[/tex] = entropy due to melting of ice at 0°C (273 K) in J/K
Δ[tex]s_{3}[/tex] = entropy due to warming liquid from 0°C (273 K) to 115°C (388 K) in J/K
Δ[tex]s_{4}[/tex] = entropy due to vaporizing liquid at 115°C (388 K) in J/K
Therefore,
Δ[tex]s_{1}[/tex] = m*[tex]c_{i}[/tex]*[tex]ln\frac{273}{263}[/tex]
m = mass = 10 g; [tex]c_{i}[/tex] = specific heat of ice = 2.106 J/(g*K)
Δ[tex]s_{1}[/tex] = 10*2.106*[tex]ln\frac{273}{263}[/tex] = 0.79 J/K
Δ[tex]s_{2}[/tex] = m*[tex]L_{f}[/tex]/273
m = 10 g; [tex]L_{f}[/tex] = 334 J/g
Δ[tex]s_{2}[/tex] = 10*334/273 = 12.23 J/K
Δ[tex]s_{3}[/tex] = m*[tex]c_{w}[/tex]*[tex]ln\frac{388}{273}[/tex]
m = mass = 10 g; [tex]c_{w}[/tex] = specific heat of water = 4.218 J/(g*K)
Δ[tex]s_{3}[/tex] = 10*4.218*[tex]ln\frac{388}{273}[/tex] = 14.83 J/K
Δ[tex]s_{4}[/tex] = m*[tex]L_{v}[/tex]/388
m = 10 g; [tex]L_{v}[/tex] = 2250 J/g
Δ[tex]s_{4}[/tex] = 10*2250/388 = 57.99 J/K
Therefore,
Δ[tex]s_{t}[/tex] = Δ[tex]s_{1}[/tex]+Δ[tex]s_{2}[/tex]+Δ[tex]s_{3}[/tex]+Δ[tex]s_{4}[/tex] = 0.79+12.23+14.83+57.99 = 85.84 J/K
Thus, the change in entropy of the system is 85.84 J/K
