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MHT CET 2026 (19 Apr, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from MHT CET 2026 (19 Apr, Shift 2)

Q8MHT CET 2026 · 19 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

1515 g of ice at 0∘0^{\circ}C is added to a vessel containing water at 40∘40^{\circ}C. The mass of water and water equivalent of the vessel is 6060 g. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be

[ LLice =80= 80 cal/g, SSwater =1= 1 cal/g ]

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Answer: (c)

  1. Heat from 60 g cooling to 0∘C0^{\circ}\text{C}: 60×40=2400 cal60 \times 40 = 2400\text{ cal}; to melt ice: 15×80=1200 cal15 \times 80 = 1200\text{ cal}, so all ice melts.
  2. 60(40−T)=1200+15T60(40 - T) = 1200 + 15T
  3. 2400−60T=1200+15T⇒T=16∘C2400 - 60T = 1200 + 15T \Rightarrow T = 16^{\circ}\text{C}
  4. So the answer is (c) 16∘C16^{\circ}\text{C}.
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