Question

The temperature of equal masses of three different liquids $\text{A, B}$ and $\text{C}$ are $10{}^{\circ }\text{C},15{}^{\circ }\text{C}$ and $20{}^{\circ }\text{C}$ respectively. The temperature when $\text{A}$ and $\text{B}$ are mixed is $13{}^{\circ }\text{C}$ and when $\text{B}$ and $\text{C}$ are mixed is $16{}^{\circ }\text{C}$. What will be the temperature when $\text{A}$ and $\text{C}$ are mixed ?

• A. $12.7{}^{\circ }\text{C}$
• B. $10.5{}^{\circ }\text{C}$
• C. $9.2{}^{\circ }\text{C}$
• D. $4.8{}^{\circ }\text{C}$

Answer 1

The correct option is A $12.7{}^{\circ }\text{C}$

Given,
$T_A=10{}^{\circ }\text{C};T_B=15{}^{\circ }\text{C};T_C=20{}^{\circ }\text{C}$
$(T_{AB}{)}_m={13}^{\circ }\text{C};(T_{BC}{)}_m={16}^{\circ }\text{C};(T_{AC}{)}_m=?$

Let, Specific heat capacities of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{A},\text{B}$ and $\text{C}$ be $S_1,S_2$ and $S_3$

When <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{A}$ and $\text{B}$ are mixed,

Applying energy conservation during process,

Energy gained by $\text{A}=$ Energy lost by $\text{B}$

$m{S}_1(13-10)=m{S}_2(15-13)$

$\Rightarrow 3S_1=2S_2........\left(1\right)$

Similarly, when $\text{B}$ and $\text{C}$ are mixed,

Energy gained by $\text{B}=$ Energy lost by $\text{C}$

$m{S}_2(16-15)=m{S}_3(20-16)$

$\Rightarrow S_2=4S_3.......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get,

$S_1=\frac{8}{3}{S}_3......\left(3\right)$

Now, when $\text{A}$ and $\text{C}$ are mixed,

Energy gained by $\text{A}=$ Energy lost by $\text{C}$

$\Rightarrow m{S}_1(T-10)=m{S}_3(20-T)$

$S_1(T-10)=S_3(20-T)$

Putting the value of $S_1$ we get,

$\left(\frac{8S_3}{3}\right)(T-10)=S_3(20-T)$

$8T-80=60-3T$

$\therefore T=\frac{140}{11}=12.7{}^{\circ }\text{C}$.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.