QJEEM20Ph8b50: JEE Main 2020 Physics - Class 11 JEEM20Ph8b50 Solution: Three containers C 1 , C 2 and C 3 have water at different…

Let containers C₁, C₂, C₃ contain water at T₁, T₂ and T₃ temperatures respectively.

Applying law of calorimetry

1(T₁ – 60) + 2(T₂ – 60) = 0

$$\Rightarrow$$ T₁ + 2T₂ = 180 ....(1)

1(T₂ – 30) + 2(T₃ – 30) = 0

$$\Rightarrow$$ T₂ + 2T₃ = 90 ......(2)

2(T₁ – 60) + 1(T₃ – 60) = 0

$$\Rightarrow$$ 2T₁ + T₃ = 180 .....(3)

from (1), (2), (3)

T₁ = 80^o C

T₂ = 50^o C

T₃ = 20^o C

1 × (T₁ – $$\theta$$) + 1 ×(T₂ – $$\theta$$) + 1(T₃ – $$\theta$$) = 0

$$\Rightarrow$$ T₁ + T₂ + T₃ = 3$$\theta$$

$$\Rightarrow$$ $$\theta$$ = $$\frac{80+50+20}{3}$$ = 50^oC