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JEE Main 2020 (3 Sep, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2020 (3 Sep, Shift 2)

Q33JEE Main 2020 · 3 SEP · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A calorimeter of water equivalent 20 g contains 180 g of water at 25oC. ‘m’ grams of steam at 100oC is mixed in it till the temperature of the mixure is 31oC. The value of ‘m’ is close to: (Latent heat of water = 540 cal g–1, specific heat of water = 1 cal g–1oC–1)

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Answer: (b)

  1. Heat gained by water + calorimeter =(180+20)×1×(31−25)=1200 cal= (180 + 20) \times 1 \times (31 - 25) = 1200\text{ cal}
  2. Heat given by steam =m(540+100−31)=609m= m(540 + 100 - 31) = 609m
  3. 609m=1200⇒m≈1.97≈2 g609m = 1200 \Rightarrow m \approx 1.97 \approx 2\text{ g}
  4. So the answer is (b) 2.
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