Question

Given three liquid having temperature ${50}^{\circ }C,{52}^{\circ }C$ and ${56}^{\circ }C$. The specific heats of the three liquid are $S,2S$ and $3S$, respectively. What will be the temperature of mixture, if each liquid is mixed in equal proportion of mass?

• A. $50.33$
• B. $53.66$
• C. $56.33$
• D. $59.66$

Answer 1

The correct option is B $53.66$

Let $m$ is mass of each liquid which are mixed and $T$ be the final temperature of the mixture.

By law of conservation of energy.

$Q_{lost}=Q_{gain}$

$\Rightarrow mS(T-50)+m\times 2S(T-52)+m\times 3S(T-56)=0$
$\Rightarrow T-50+2T-104+3T-168=0$
$\Rightarrow 6T=322$
$T={53.66}^{\circ }\text{C}$

Hence, $\left(B\right)$ is the correct answer.

Why this question? This question gives you clear concept of mixing of liquid and their dependencies on specific heat.