Two liquids A and B are at temperatures 32∘C and 24∘C respectively. When mixed in equal masses the temperature of the mixture is found to be 28∘C. The ratio of their specific heat is:
A
1:1
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B
4:3
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C
2:3
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D
3:2
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Solution
The correct option is A
1:1
Given, temperature of liquid A, TA=32∘C, temperature of liquid B, TB=24∘C and temperature of the mixture, T=28∘C
Let where, mA and mB be the mass of body A and B, CA and CB be the specific heat of body A and B, then, Heat lost by A=Heat gained by B
or, mA×CA×(TA−T)=mB×CB×(T−TB)
Since mA=mB, we have, CA×(32−28)=CB×(28−24)⇒CACB=11
Therefore the ratio of specific heat is 1:1.