Question

Two liquids A and B are at $30{}^\circ C$ and $20{}^\circ C$ respectively. When they are mixed in equal masses, the temperature of the mixture is found to be $26{}^\circ C$. The ratio of specific heat is
A. 4:3
B. 3:4
C. 2:3
D. 3:2

Answer 1

Hint: As a first step, you could recall the expression for specific heat. Now you could rearrange and get the heat lost and gained by each liquid. On mixing, we know that higher temperature liquid loss heat and lower temperature gain heat to achieve an equilibrium temperature. Thus you could equate the heat lost and gained by both liquids, rearrange and substitute to get the required ratio.

Formula used:
Expression for specific heat,
$C=\dfrac{Q}{m\Delta T}$

Complete step by step solution:
In the question we are given two liquids A and B that are at different temperatures $30{}^\circ C$ and $20{}^\circ C$ respectively. We are now mixing equal mass of both liquids and the mixture is found to be$26{}^\circ C$. From these given information, we have to find the ratio of specific heats of these liquids.
Firstly, let us recall what specific heat of a liquid is.
The specific heat capacity of a substance is the amount of heat required by 1Kg of a substance to result in unit degree change (rise or fall) in temperature. It can be expressed as,
$C=\dfrac{Q}{m\Delta T}$
$\Rightarrow Q=mC\Delta T$ ……………………………. (1)
Where, Q is the heat supplied, ΔT is the resultant temperature change and m is the given amount of substance.
But we know that when the two liquids at different temperatures are mixed then, the heat lost by one liquid will be equal to the heat gained by the other.
$-{{Q}_{A}}={{Q}_{B}}$ ………………………………… (2)
Using (1),
$-{{m}_{A}}{{C}_{A}}\Delta {{T}_{A}}={{m}_{B}}{{C}_{B}}\Delta {{T}_{B}}$
But the amount of liquid is the same m.
$\Rightarrow -{{C}_{A}}\left( T-{{T}_{A}} \right)={{C}_{B}}\left( T-{{T}_{B}} \right)$
Substituting the values,
$-{{C}_{A}}\left( 26-30 \right)={{C}_{B}}\left( 26-20 \right)$
$\Rightarrow \dfrac{{{C}_{A}}}{{{C}_{B}}}=\dfrac{6}{4}$
$\therefore {{C}_{A}}:{{C}_{B}}=3:2$
Therefore, we found that the ratio of specific heats of A and B is 3:2.

Hence, option D is the correct answer.

Note:
You may have noticed the negative sign in equation (2). This negative sign arises due to the fact that one liquid is losing heat while the other gains heat. The loss and gain of heat is the resultant of the increase and decrease of temperature which happens due to the mixing of the two liquids at different temperatures.