JEE Main · 2025 · Shift-IhardTHERMO-150

Total enthalpy change for freezing of 1 mol of water at 10°C to ice at -10°C is _____. Given: fusH = x\ kJ/mol…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Total enthalpy change for freezing of 1 mol of water at 10°C10°\text{C} to ice at −10°C-10°\text{C} is _____.

Given: ΔfusH=x kJ/mol\Delta_{\text{fus}}H = x\ \text{kJ/mol}

Cp[\ceH2O(l)]=y J mol−1K−1C_p[\ce{H2O(l)}] = y\ \text{J mol}^{-1}\text{K}^{-1}

Cp[\ceH2O(s)]=z J mol−1K−1C_p[\ce{H2O(s)}] = z\ \text{J mol}^{-1}\text{K}^{-1}

Options
  1. a

    −x−10y−10z-x - 10y - 10z

  2. b

    −10(100x+y+z)-10(100x + y + z)

    ✓
  3. c

    10(100x+y+z)10(100x + y + z)

  4. d

    x−10y−10zx - 10y - 10z

Correct Answerb

−10(100x+y+z)-10(100x + y + z)

Detailed Solution

🧠 Three-step cooling: liquid 10°C → 0°C, then freeze, then cool ice 0°C → −10°C Each step contributes differently: heating steps use CPΔTC_P\Delta T, phase change uses ΔHfus\Delta H_{\text{fus}}.

🗺️ Energy balance (signs: heat released = negative) Step 1 — cool liquid water, 10→0 ∘C10 \rightarrow 0\,^\circ\text{C} (ΔT=−10 K\Delta T = -10\,\text{K}): ΔH1=1 mol×y J K−1mol−1×(−10)=−10y J\Delta H_1 = 1\,\text{mol} \times y\,\text{J K}^{-1}\text{mol}^{-1} \times (-10) = -10y\,\text{J}

Step 2 — freeze at 0°C (reverse of fusion, so negative): ΔH2=−x kJ=−1000x J\Delta H_2 = -x\,\text{kJ} = -1000x\,\text{J}

Step 3 — cool ice, 0→−10 ∘C0 \rightarrow -10\,^\circ\text{C}: ΔH3=1 mol×z×(−10)=−10z J\Delta H_3 = 1\,\text{mol} \times z \times (-10) = -10z\,\text{J}

Total =−1000x−10y−10z=−10(100x+y+z)= -1000x - 10y - 10z = -10(100x + y + z)

Answer: (b) −10(100x+y+z)\boxed{\text{Answer: (b) } -10(100x + y + z)}

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