Cooling + Freezing Energy Change ⚡ JEE Shortcut
❓ Question Find the total enthalpy change when 1 mol of water is converted from: 10 ∘ C → − 10 ∘ C (ice) 10^\circ C \rightarrow -10^\circ C \text{ (ice)} 🖼 Question Image ✍️ Short Concept This process happens in three steps : 1️⃣ Cool water from 10°C → 0°C 2️⃣ Freeze water at 0°C 3️⃣ Cool ice from 0°C → −10°C Total enthalpy change = sum of all three. 🔷 Step 1 — Cooling Water (10°C → 0°C) 💯 Heat released: q 1 = n C w a t e r Δ T q_1 = n C_{water} \Delta T For water: C w a t e r = 75 J m o l − 1 K − 1 C_{water} = 75 \, J\, mol^{-1} K^{-1} q 1 = 1 × 75 × ( 10 ) q_1 = 1 \times 75 \times (10) q 1 = 750 J q_1 = 750 \, J 🔷 Step 2 — Freezing at 0°C Latent heat of fusion: Δ H f = 6.01 k J / m o l \Delta H_f = 6.01 \, kJ/mol Heat released: q 2 = 6.01 k J q_2 = 6.01 \, kJ 🔷 Step 3 — Cooling Ice (0°C → −10°C) Heat released: q 3 = n C i c e Δ T q_3 = n C_{ice} \Delta T For ice: C i c e = 37.6 J m o l − 1 K − 1 C_{ice} = 37.6 \, J\, mol^{-1}K^{-1} q ...