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JEE Main 2013 (23 Apr, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2013 (23 Apr, Shift 1)

Q50JEE Main 2013 · 23 APR · SHIFT 1Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

500 g of water and 100 g of ice at 0°C are in a calorimeter whose water equivalent is 40 g. 10 g of steam at 100°C is added to it. Then water in the calorimeter is: (Latent heat of ice = 80 cal/g, Latent heat of steam = 540 cal/g)

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Answer: (b)

  1. Heat steam can give down to 0∘0^{\circ}C: 10(540+100)=6400 cal10(540 + 100) = 6400\ \text{cal}.
  2. Heat to melt all ice: 100×80=8000 cal>6400100 \times 80 = 8000\ \text{cal} > 6400, so final temperature is 0∘0^{\circ}C.
  3. Ice melted =640080=80 g= \dfrac{6400}{80} = 80\ \text{g}
  4. Water =500+80+10=590 g= 500 + 80 + 10 = 590\ \text{g}
  5. So the answer is (b) 590 g.
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