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MHT CET 2025 (23 Apr, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from MHT CET 2025 (23 Apr, Shift 2)

Q15MHT CET 2025 · 23 APR · SHIFT 2Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A stationary object at 4∘C4^{\circ}\text{C} and weighing 3.5 kg falls from a height of 2000 m on snow mountain at 0∘C0^{\circ}\text{C}. If the temperature of the object just before hitting the snow is 0∘C0^{\circ}\text{C} and the object comes to rest immediately then the quantity of ice that melts is (Acceleration due to gravity =10 m/s2= 10\text{ m/s}^{2}, Latent heat of ice =3.5×105 J/kg= 3.5 \times 10^{5}\text{ J/kg})

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Answer: (c)

  1. The object stops at 0∘0^{\circ}C, so its mechanical energy mghmgh goes into melting ice: mgh=miceLmgh = m_{ice}L.
  2. mgh=3.5×10×2000=7×104 Jmgh = 3.5 \times 10 \times 2000 = 7 \times 10^{4}\ \text{J}
  3. mice=7×1043.5×105=0.2 kgm_{ice} = \dfrac{7 \times 10^{4}}{3.5 \times 10^{5}} = 0.2\ \text{kg}
  4. So the answer is (c) 200 gram.
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