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JEE Main 2019 (11 Jan, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2019 (11 Jan, Shift 2)

Q39JEE Main 2019 · 11 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

When 100 g of a liquid A at 100oC is added to 50 g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100 g of liquid A at 100oC is added to 50 g of liquid B at 50oC, will be:

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Answer: (d)

  1. First mixing: 100 sA(100−90)=50 sB(90−75)⇒sB=43sA100\,s_A(100 - 90) = 50\,s_B(90 - 75) \Rightarrow s_B = \tfrac{4}{3}s_A
  2. Second mixing: 100 sA(100−T)=50⋅43sA(T−50)100\,s_A(100 - T) = 50 \cdot \tfrac{4}{3}s_A(T - 50)
  3. 300(100−T)=200(T−50)⇒T=80∘C300(100 - T) = 200(T - 50) \Rightarrow T = 80^{\circ}\text{C}
  4. So the answer is (d) 80∘C80^{\circ}\text{C}.
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