Question

46. Two liquids A and B are at 32o C and 24o C. When mixed in equal masses the temperature of the mixture is found

to be 28o C. Their sp. heats

are in the ratio of

(A) 3 : 2 (B) 2 : 3

(C) 1 : 1 (D) 4 : 3​

Answer 1

Answer:

Explanation:

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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Answer 2

Answer:

Explanation:

given that temperature of A = 32 degrees

the temperature of B = 24 degrees

at equilibrium temperatures of both A and B is 28 degrees

let mass of a = mass of B =m

heat lost by A = heat gained by B

mSa(t1-t2)=mSb(t2-t1)

Sa(32-28)=Sb(28-24)

Sa(4)=Sb(4)

Sa:Sb=1:1

therefore the specific heats of A and B are in the ratio 1:1