Two samples, each containing 400g of water, are mixed. The temperature of sample A before mixing was 20∘C and that of sample B was 65∘C. What will be the final temperature of the mixture? (Specific heat of water = 4186 J/kg K)
A
22.5∘C
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B
42.5∘C
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C
32.5∘C
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D
52.5∘C
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Solution
The correct option is B42.5∘C Given: Mass of sample A and B, mA=mB=400g=0.4kg Temperature of sample A, TA=20∘C Temperature of sample B, TB=65∘C Specific heat of water, Sw=4186J/kgK The heat lost by sample A will be equal to heat gained by sample B. Let the temperature of the mixture be Tm. Heat lost = Heat gained ∴mASw(TB−Tm)=mBSw(Tm−TA) ⇒0.4×(65−Tm)=0.4×(Tm−20) ⇒2Tm=85 ⇒Tm=42.5∘C