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For freezing, Δ H = -n × Δ Hfus = -3 × 5.5 = -16.5 kJ, Δ Ssurr = -Δ H / T = -(-16.5 × 10³) / 250 = 66 J/K.
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy
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