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Calorimetry

Two liquids A...

Question

Two liquids A and B are at $32^{\circ} C$ and $24^{\circ} C .$ When mixed in equal masses the temperature of the mixture is
found to be $28^{\circ} C .$ Their specific heats are in the ratio of

3 : 2

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2 : 3

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1 : 1

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4 : 3

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Solution

The correct option is C 1 : 1
Heat lost by A = Heat gained by B
$\Rightarrow m_{A} \times C_{A} \times (T_{A} - T) = m_{B} \times C_{B} \times (T - T_{B}) S i n c e m_{A} = m_{B} a n d T e m p e r a t u r e o f t h e m i x t u r e (T) = 28^{\circ} C ∴ C_{A} \times (32 - 28) = C_{B} \times (28 - 24) \Rightarrow \frac{C_{A}}{C_{B}} = 1 : 1$

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Similar questions

Two liquids $A$ and $B$ are at temperatures $32^{\circ} C$ and $24^{\circ} C$ respectively. When mixed in equal masses the temperature of the mixture is found to be $28^{\circ} C$ . The ratio of their specific heat is:

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

The correct option is C 1 : 1Heat 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

Two liquids A and B are at $32^{\circ} C$ and $24^{\circ} C .$ When mixed in equal masses the temperature of the mixture is
found to be $28^{\circ} C .$ Their specific heats are in the ratio of