Cooling + Freezing Energy Change ⚡ JEE Shortcut

❓ Question

Find the total enthalpy change when 1 mol of water is converted from:

$$10^\circ C \rightarrow -10^\circ C \text{ (ice)}$$

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🖼 Question Image

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✍️ 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.

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🔷 Step 1 — Cooling Water (10°C → 0°C) 💯

Heat released:

$$q_1 = n C_{water} \Delta T$$

For water:

$$C_{water} = 75 \, J\, mol^{-1} K^{-1}$$
$$q_1 = 1 \times 75 \times (10)$$
$$q_1 = 750 \, J$$

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🔷 Step 2 — Freezing at 0°C

Latent heat of fusion:

$$\Delta H_f = 6.01 \, kJ/mol$$

Heat released:

$$q_2 = 6.01 \, kJ$$

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🔷 Step 3 — Cooling Ice (0°C → −10°C)

Heat released:

$$q_3 = n C_{ice} \Delta T$$

For ice:

$$C_{ice} = 37.6 \, J\, mol^{-1}K^{-1}$$
$$q_3 = 1 \times 37.6 \times 10$$
$$q_3 = 376 \, J$$

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🔷 Step 4 — Total Enthalpy Change

Convert everything to kJ:

$$q_1 = 0.75 \, kJ$$
$$q_3 = 0.376 \, kJ$$

Total heat released:

$$q = 0.75 + 6.01 + 0.376$$
$$q = 7.136 \, kJ$$

Since heat is released:

$$\Delta H = -7.14 \, kJ$$

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✅ Final Answer

$$\boxed{\Delta H \approx -7.14 \, kJ}$$

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⭐ Golden JEE Insight

Whenever temperature crosses phase change:

Always split into three steps:

1️⃣ Cooling
2️⃣ Phase change
3️⃣ Cooling again

Never try solving in one formula.