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JEE Main 2020 (8 Jan, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2020 (8 Jan, Shift 2)

Q34JEE Main 2020 · 8 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Three containers C1, C2 and C3 have water at different temperatures. The table below shows the final temperature T when different amounts of water (given in litres) are taken from each containers and mixed (assume no loss of heat during the process)

Figure

The value of θ\theta (in °C to the nearest integer) is ..........

Numerical answer type. Work it out, then open the answer.

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Answer: 50

  1. For mixing equal-specific-heat water: T=∑ViTi∑ViT = \dfrac{\sum V_i T_i}{\sum V_i}.
  2. Row 1: T1+2T2=180T_1 + 2T_2 = 180; Row 2: T2+2T3=90T_2 + 2T_3 = 90; Row 3: 2T1+T3=1802T_1 + T_3 = 180
  3. Adding: 3(T1+T2+T3)=450⇒T1+T2+T3=1503(T_1+T_2+T_3) = 450 \Rightarrow T_1+T_2+T_3 = 150
  4. θ=T1+T2+T33=50∘C\theta = \dfrac{T_1+T_2+T_3}{3} = 50^{\circ}\text{C}
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