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JEE Main 2019 (8 Apr, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2019 (8 Apr, Shift 1)

Q43JEE Main 2019 · 8 APR · SHIFT 1Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A thermally insulated vessel contains 150g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water = 2.10 × 106 J kg–1 and Latent heat of Fusion of water = 3.36 × 105 J kg–1)

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Answer: (c)

  1. Heat for evaporation comes from water freezing: mvLv=miLfm_v L_v = m_i L_f, with mv+mi=150 gm_v + m_i = 150\ \text{g}.
  2. mv=150×LfLf+Lv=150×3.36×1053.36×105+2.10×106m_v = \dfrac{150 \times L_f}{L_f + L_v} = \dfrac{150 \times 3.36 \times 10^{5}}{3.36 \times 10^{5} + 2.10 \times 10^{6}}
  3. mv=150×0.3362.436≈20.7 gm_v = \dfrac{150 \times 0.336}{2.436} \approx 20.7\ \text{g}
  4. So the answer is (c) 20 g.
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