Question

Two liquids $A$ and $B$ are at temperatures ${32}^{\circ }C$ and ${24}^{\circ }C$ respectively. When mixed in equal masses the temperature of the mixture is found to be ${28}^{\circ }C$. The ratio of their specific heat is:

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

Answer 1

The correct option is C 1 : 1

Given, temperature of liquid A, $T_A={32}^{\circ }C$, temperature of liquid B, $T_B={24}^{\circ }C$ and temperature of the mixture, $T={28}^{\circ }C$
Let where, $m_A$ and $m_B$ be the mass of body A and B, $C_A$ and $C_B$ be the specific heat of body A and B, then,
$\text{Heat lost by A}=\text{Heat gained by B}$
or, $m_A\times C_A\times (T_A-T)=m_B\times C_B\times (T-T_B)$

Since $m_A=m_B$, we have,
$C_A\times (32-28)=C_B\times (28-24)$ $\Rightarrow \frac{C_A}{C_B}=\frac{1}{1}$
Therefore the ratio of specific heat is 1:1.