Question

A thermally insulated vessel contains two liquids at temperature $T_1$ and $T_2$ respectively and specific heats $C_1$ and $C_2$ separated by a non- conducting partition. The partition is removed and the difference between the initial temperature of either of the liquids and the temperature T established in the vessel turns out to be equal to half the difference between the initial temperature of the liquids. Determine the ratio $\frac{m_1}{m_2}$ (masses of the liquids).

• A. $\frac{C_1}{C_2}$
• B. $\frac{C_2}{C_1}$
• C. $\frac{C_2+C_1}{C_2-C_1}$
• D. $\frac{C_2-C_1}{C_2+C_1}$

Answer 1

The correct option is B $\frac{C_2}{C_1}$

$T-T_2=T_1-T=\frac{(T_1-T_2)}{2}$ (Given T is temperature of the mixture)
Appply heat lost = heat gained
$m_1{C}_1(T_1-T)=m_2{C}_2(T-T_2)$
$\frac{m_1}{m_2}=\frac{C_2(T-T_2)}{C_1(T_1-T)}=\frac{C_2}{C_1}$