0.1 m3 of water at 80°C is mixed with 0.3 m3 of water at 60°C. The final temperature of the mixture is:

Heat lost = Heat gained ⇒ (0.1)ρ×s×(80-θ)=(0.3)ρ×s×(θ-60)⇒θ=65oC

0.1 m3 of water at 80°C is mixed with 0.3 m3 of water at 60°C. The final temperature of the mixture is:

Heat lost = Heat gained ⇒ (0.1)ρ×s×(80-θ)=(0.3)ρ×s×(θ-60)⇒θ=65oC

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0.1 m³ of water at 80°C is mixed with 0.3 m³ of water at 60°C. The final temperature of the mixture is:

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60°C

65°C

75°C

70°C

answer is A.

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Detailed Solution

Heat lost = Heat gained
$\Rightarrow (0.1) ρ \times s \times (80 - θ) = (0.3) ρ \times s \times (θ - 60) \Rightarrow θ = 65^{o} C$

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0.1 m3 of water at 80°C is mixed with 0.3 m3 of water at 60°C. The final temperature of the mixture is:

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0.1 m³ of water at 80°C is mixed with 0.3 m³ of water at 60°C. The final temperature of the mixture is: