Question

Two liquids A and B are at $32^\circ {\text{C}}$ and $24^\circ {\text{C}}$. When missed in equal masses the temperature of the mixture is found to be $28^\circ {\text{C}}$. Their specific heats are in the ratio of
A. $3:2$
B. $2:3$
C. $1:1$
D. $4:3$

Answer 1

Use the formula for specific heat of a substance. This formula gives the relation between the amount of the heat exchanged by the substance, mass of the substance and the change in temperature of the substance. Equate the heats exchanged by the liquids A and B and determine the ratio of the specific heats.

Formula used:
The expression for specific heat $C$ of a substance is given by
$C = \dfrac{Q}{{m\left( {{T_f} - {T_i}} \right)}}$....................(1)
Here, $Q$ is the amount of heat exchanged, $m$ is the mass of the substance, ${T_i}$ is the initial temperature of the substance and ${T_f}$ is the final temperature of the substance.

Complete step by step answer:
We have given that the initial temperatures of the two liquids A and B are $32^\circ {\text{C}}$ and $24^\circ {\text{C}}$ respectively.
${T_A} = 32^\circ {\text{C}}$
${T_B} = 24^\circ {\text{C}}$
The final temperature of the mixture of the liquids A and B is $28^\circ {\text{C}}$.
${T_f} = 28^\circ {\text{C}}$
Let $m$ be the mass of the two liquids A and B mixed.
Rewrite equation (1) for the specific heat ${C_A}$ of the liquid A.
${C_A} = \dfrac{{{Q_A}}}{{m\left( {{T_A} - {T_f}} \right)}}$
Here, ${Q_A}$ is the amount of heat released by liquid A.
Rearrange the above equation for ${Q_A}$.
${Q_A} = {C_A}m\left( {{T_A} - {T_f}} \right)$
Rewrite equation (1) for the specific heat ${C_B}$ of the liquid B.
${C_B} = \dfrac{{{Q_B}}}{{m\left( {{T_f} - {T_B}} \right)}}$
Here, ${Q_B}$ is the amount of heat absorbed by liquid B.
Rearrange the above equation for ${Q_B}$.
${Q_B} = {C_B}m\left( {{T_f} - {T_B}} \right)$
The amount of heat released by liquid A is equal to the amount of heat absorbed by liquid B.
${Q_A} = {Q_B}$
Substitute ${C_A}m\left( {{T_A} - {T_f}} \right)$ for ${Q_A}$ and ${C_B}m\left( {{T_f} - {T_B}} \right)$ for ${Q_B}$ in the above equation.
${C_A}m\left( {{T_A} - {T_f}} \right) = {C_B}m\left( {{T_f} - {T_B}} \right)$
$\Rightarrow {C_A}\left( {{T_A} - {T_f}} \right) = {C_B}\left( {{T_f} - {T_B}} \right)$
Rearrange the above equation for $\dfrac{{{C_A}}}{{{C_B}}}$.
$\Rightarrow \dfrac{{{C_A}}}{{{C_B}}} = \dfrac{{{T_f} - {T_B}}}{{{T_A} - {T_f}}}$
Substitute $28^\circ {\text{C}}$ for ${T_f}$, $32^\circ {\text{C}}$ for ${T_A}$ and $24^\circ {\text{C}}$ for ${T_B}$ in the above equation.
$\Rightarrow \dfrac{{{C_A}}}{{{C_B}}} = \dfrac{{28^\circ {\text{C}} - 24^\circ {\text{C}}}}{{32^\circ {\text{C}} - 28^\circ {\text{C}}}}$
$\Rightarrow \dfrac{{{C_A}}}{{{C_B}}} = \dfrac{4}{4}$
$\therefore \dfrac{{{C_A}}}{{{C_B}}} = \dfrac{1}{1}$

Therefore, the ratio of the specific heats of liquids A and B is 1:1. Hence, the correct option is C.

Note: The students should be careful while writing the formula for specific heat of the liquids A and B as the initial temperature of the liquid A is more than the temperature of the mixture and initial temperature of the liquid B is less than the temperature of the mixture. Otherwise one may get a result as the negative ratio of the specific heat of the liquids.