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JEE Main 2025 (22 Jan, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2025 (22 Jan, Shift 1)

Q14JEE Main 2025 · 22 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

An amount of ice of mass 10−3 kg10^{-3}\text{ kg} and temperature −10∘C-10^{\circ}\text{C} is transformed to vapour of temperature 110∘C110^{\circ}\text{C} by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice =2100 J kg−1 K−1= 2100\text{ J kg}^{-1}\text{ K}^{-1}, specific heat of water =4180 J kg−1 K−1= 4180\text{ J kg}^{-1}\text{ K}^{-1}, specific heat of steam =1920 J kg−1 K−1= 1920\text{ J kg}^{-1}\text{ K}^{-1}, Latent heat of ice =3.35×105 J kg−1= 3.35 \times 10^{5}\text{ J kg}^{-1} and Latent heat of steam =2.25×106 J kg−1= 2.25 \times 10^{6}\text{ J kg}^{-1})

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Answer: (b)

  1. Ice −10→0∘C-10 \to 0^{\circ}\text{C}: 10−3×2100×10=21 J10^{-3} \times 2100 \times 10 = 21\text{ J}; melting: 335 J335\text{ J}.
  2. Water 0→100∘C0 \to 100^{\circ}\text{C}: 10−3×4180×100=418 J10^{-3} \times 4180 \times 100 = 418\text{ J}; vaporising: 2250 J2250\text{ J}.
  3. Steam 100→110∘C100 \to 110^{\circ}\text{C}: 10−3×1920×10=19.2 J10^{-3} \times 1920 \times 10 = 19.2\text{ J}
  4. Total =21+335+418+2250+19.2≈3043 J= 21 + 335 + 418 + 2250 + 19.2 \approx 3043\text{ J}
  5. So the answer is (b) 3043 J3043\text{ J}.
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