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Moles of ice = 36 / 18 = 2 mol. Heat absorbed = 2 × 6.01 = 12.02 kJ.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
For freezing, Δ H = -n × Δ Hfus = -2 × 6.01 = -12.02 kJ, Δ Ssurr = -Δ H / T = -(-12.02 × 10³) / 273 = 44.03 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems
Moles = 36 / 18 = 2 mol, Δ S = n × Δ Hfus / T = 2 × 6.01 × 10³ / 273 = 44 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium
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