Cooling + Freezing Combo ⚡ Thermodynamics JEE

 

❓ Concept Question

How do we calculate total enthalpy change when water is cooled, frozen, and further cooled as ice?

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🖼 Concept Image

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✍️ Short Concept

This is a multi-step thermodynamics process.

👉 Never solve in one step
👉 Always break into 3 parts

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🔷 Step 1 — Process is NOT Single Step 💯

Sequence:

Water (10°C) → Water (0°C) → Ice (0°C) → Ice (−10°C)

👉 Total = 3 energy changes

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🔷 Step 2 — Cooling of Liquid Water

$$\Delta H_1 = m C_p(\text{liquid}) \Delta T$$

From:

$$10^\circ C \rightarrow 0^\circ C$$

Heat is released:

$$\Delta H_1 < 0$$

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🔷 Step 3 — Freezing at 0°C

$$\Delta H_2 = - \Delta H_{\text{fusion}}$$

👉 Freezing releases latent heat

No temperature change, only phase change.

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🔷 Step 4 — Cooling of Ice

$$\Delta H_3 = m C_p(\text{solid}) \Delta T$$

From:

$$0^\circ C \rightarrow -10^\circ C$$

Again heat released:

$$\Delta H_3 < 0$$

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🔷 Step 5 — Golden JEE Rule

Total enthalpy:

$$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3$$

👉 Add all three contributions

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✅ Final Takeaway

$$\boxed{\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3}$$

Since all steps release heat:

$$\boxed{\Delta H \text{ is NEGATIVE}}$$

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⭐ Golden JEE Insight

Whenever:

👉 Temperature crosses phase change point

Always:

1️⃣ Cool
2️⃣ Change phase
3️⃣ Cool again

👉 Never skip steps.