JEE Main · 2019 · Shift-IIhardTHERMO-129

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is ______.…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is ______. (Specific heat of water liquid and water vapour are 4.2 kJ K−14.2\,\text{kJ K}^{-1} and 2.0 kJ K−1kg−12.0\,\text{kJ K}^{-1}\text{kg}^{-1}; heat of liquid fusion and vaporization of water are 334 kJ kg−1334\,\text{kJ kg}^{-1} and 2491 kJ kg−12491\,\text{kJ kg}^{-1}, respectively). (log⁡273=2.436\log 273 = 2.436, log⁡373=2.572\log 373 = 2.572, log⁡383=2.583\log 383 = 2.583)

Options
  1. a

    9.26 kJ kg−1K−19.26\,\text{kJ kg}^{-1}\text{K}^{-1}

    ✓
  2. b

    2.64 kJ kg−1K−12.64\,\text{kJ kg}^{-1}\text{K}^{-1}

  3. c

    8.49 kJ kg−1K−18.49\,\text{kJ kg}^{-1}\text{K}^{-1}

  4. d

    7.90 kJ kg−1K−17.90\,\text{kJ kg}^{-1}\text{K}^{-1}

Correct Answera

9.26 kJ kg−1K−19.26\,\text{kJ kg}^{-1}\text{K}^{-1}

Detailed Solution

🧠 Multi-stage entropy: sum ΔS=qrev/T\Delta S = q_{\text{rev}}/T for phase changes and nCPln⁡(T2/T1)n C_P \ln(T_2/T_1) for heating Ice at 273 K → water at 373 K → steam at 383 K involves four separate entropy contributions that simply add.

🗺️ Calculation (per kg) Stage 1 — Fusion at 273 K: ΔS1=334273=1.223 kJ K−1kg−1\Delta S_1 = \frac{334}{273} = 1.223\,\text{kJ K}^{-1}\text{kg}^{-1}

Stage 2 — Heating liquid 273 → 373 K: ΔS2=4.2×2.303×(log⁡373−log⁡273)=4.2×2.303×0.136=1.315\Delta S_2 = 4.2 \times 2.303 \times (\log 373 - \log 273) = 4.2 \times 2.303 \times 0.136 = 1.315

Stage 3 — Vaporization at 373 K: ΔS3=2491373=6.678\Delta S_3 = \frac{2491}{373} = 6.678

Stage 4 — Heating steam 373 → 383 K: ΔS4=2.0×2.303×(log⁡383−log⁡373)=2.0×2.303×0.011=0.051\Delta S_4 = 2.0 \times 2.303 \times (\log 383 - \log 373) = 2.0 \times 2.303 \times 0.011 = 0.051

ΔStotal=1.223+1.315+6.678+0.051=9.267 kJ K−1kg−1\Delta S_{\text{total}} = 1.223 + 1.315 + 6.678 + 0.051 = 9.267\,\text{kJ K}^{-1}\text{kg}^{-1}

Answer: (a) 9.26 kJ kg−1K−1\boxed{\text{Answer: (a) } 9.26\,\text{kJ kg}^{-1}\text{K}^{-1}}

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