Question

Equal masses of three liquids A, B and C have temperatures ${10}^{\circ }C$, ${25}^{\circ }C$ and ${40}^{\circ }C$ respectively. If A and B are mixed, the mixture has a temperature of 15°C. If B and C are mixed the mixture has a temperature of ${30}^{\circ }C$. If A and C are mixed, the mixture will have a temperature of

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We know that,

Heat lost by a liquid = Heat gained by another liquid

For liquids A and B we have,
$m{S}_A(15-10)=m{S}_B(25-10)$
$\Rightarrow S_A=2S_B\dots \dots \left(i\right)$

Similarly for liquids B and C we have,
$m{S}_B(30-25)=m{S}_C(40-30)$
$\Rightarrow S_B=2S_C\dots \dots \left(ii\right)$

Now for liquids A and C we have,
$m{S}_A(T-10)=m{S}_C(40-T)\dots \dots \left(iii\right)$
putting values from equations (i) and (ii) in equation (ii) we get,
$2S_B(T-10)=0.5S_B(40-T)$
$\therefore T={16}^{\circ }C$.

Thus if A and C are mixed, the mixture will have a temperature of ${16}^{\circ }C$.