When 100 g of a liquid A at 10 0 ∘ C is added to 50 g of a liquid B at temperature 7 5 ∘ C , the temperature of the mixture becomes 9 0 ∘ C . The temperature of the mixture, if 100 g of liquid A at 10

Question

When 100 g of a liquid A at 10 0 ∘ C is added to 50 g of a liquid B at temperature 7 5 ∘ C , the temperature of the mixture becomes 9 0 ∘ C . The temperature of the mixture, if 100 g of liquid A at 10 0 ∘ C is added to 50 g of liquid B at 5 0 ∘ C , will be:

TestHub · Accepted Answer

Heat loss = Heat gain = mS Δ θ So , m A ​ S A ​ Δ θ A ​ = m B ​ S B ​ Δ θ B ​ ⇒ 100 × S A ​ × ( 100 − 90 ) = 50 × S B ​ × ( 90 − 75 ) 2 S A ​ = 1.5 S B ​ ⇒ S A ​ = 4 3 ​ S B ​ ​ Now, 100 × S A ​ × ( 100 − θ ) = 50 × S B ​ × ( θ − 50 ) 2 × ( 4 3 ​ ) × ( 100 − θ ) = ( θ − 50 ) 300 − 3 θ = 2 θ − 100 400 = 5 θ ⇒ θ = 8 0 ∘ C ​

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