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NEET 1988 (AIPMT) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from NEET 1988 (AIPMT)

Q56NEET 1988 · AIPMTMedium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

10 gm of ice cubes at 0∘C0^{\circ}\text{C} are released in a tumbler (water equivalent 55 g) at 40∘C40^{\circ}\text{C}. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L=80 cal/gL = 80\text{ cal/g})

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Answer: (b)

  1. Heat lost by tumbler = heat to melt ice + warm melt water.
  2. 55(40−T)=10×80+10T55(40 - T) = 10 \times 80 + 10T
  3. 2200−55T=800+10T⇒T=140065≈21.5∘C2200 - 55T = 800 + 10T \Rightarrow T = \dfrac{1400}{65} \approx 21.5^{\circ}\text{C}
  4. So the answer is (b) 22∘C22^{\circ}\text{C}.
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