Question

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

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

Answer 1

The correct option is C 1 : 1

Heat lost by A = Heat gained by B
$\Rightarrow m_A\times C_A\times (T_A-T)=m_B\times C_B\times (T-T_B)Since{m}_A=m_{B}andTemperatureofthemixture\left(T\right)={28}^{\circ }C\therefore C_A\times (32-28)=C_B\times (28-24)\Rightarrow \frac{C_A}{C_B}=1:1$