Question

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

• A. $3:2$
• B. $2:3$
• C. $1:1$
• D. $4:3$

Answer 1

The correct option is C $1:1$

Let specific heats of $\text{A}$ and $\text{B}$ be $s_A$ and $s_B$ respectively.

Also, their masses are $m$ and final temperature is ${28}^{\circ }\text{C}.$

Heat released by $\text{A}=$ Heat absorbed by $\text{B}$

$\Rightarrow m{s}_A({32}^{\circ }\text{C}-{28}^{\circ }\text{C})=m{s}_B({28}^{\circ }\text{C}-{24}^{\circ }\text{C})$

$\Rightarrow \frac{s_A}{s_B}=\frac{4}{4}=1:1$

Thus, the required ratio is $1:1$.