10 g of ice at 0° are added to 20 g water at 90° in a thermally insulated flask of Negligible heat capacity. The heat of fusion of ice is 6 KJ/mol, change in entropy for system ..? Cp = 75.42 KJ/mol Pls answer fast only final answer is also ok
Answers
ΔS = 58.37 J/K
Explanation:
Heat gained by ice = heat lost by warm water.
Energy required to melt the ice Q1 = m*dH fus = 10 * 6 KJ/mol = 60000 J
Energy required to warm the melted ice Q2 = m *Cp *dT
dT = (Tf -0°C )
Q2 = 10 * 75.42 * (Tf -0°C )
Energy to cool the liquid water Q3 = m * Cp * dT = 20 * 75.42 J/mol * (900C-Tf)
So Q1 + Q2 = Q3
60000 + 10 * 75.42 * (Tf -0°C ) = 20 * 75.42 * (900C-Tf)
Solving, we get Tf = 34.27°C
Entropy change dS = dSi + dSw
Entropy change of ice:
ΔSi = mi (H fus / T fus + C. ln. T final/T us )
= 10 ( 6000/273 + 75.42 x 0.118
= 10x. 8429.5/ 273
ΔSi = 308.77 J/K
Entropy change of water:
ΔSw = mw. C. ln. T final /T water
= 20. x. 75.42x (-0.166)
ΔSw = -250.4 J/K
Therefore ΔS = ΔSi + ΔSw
= 308.77 + (−250.4)
= 58.37 J/K