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Q.

Two liquids A and B are at 32oC and 24oC. When mixed in equal masses the temperature of the mixture is found to be 28oC. Their specific heats are in the ratio of

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a

3:2

b

2:3

c

1:1

d

4:3

answer is C.

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Detailed Solution

Heat lost by A= Heat gained by B⇒ mA×cA×TA−T=mB×cB×T−TB Since mA=mB and temperature of the mixture (T)=28∘C∴ cA×(32−28)=cB×(28−24)⇒ cAcB=1:1
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