Question

Two liquids of the same mass which are at temperatures ${20}^{\circ }\text{C}$ and ${40}^{\circ }\text{C}$ are mixed to form a composite liquid. If the temperature of the mixture is ${32}^{\circ }\text{C}$, then the ratio of their heat capacities is

• A. $\frac{1}{3}$
• B. $\frac{4}{9}$
• C. $\frac{9}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Given,
Initial temperature of liquid 1 $\left({\theta }_1\right)={20}^{\circ }\text{C}$
Initial temperature of liquid 2 $\left({\theta }_2\right)={40}^{\circ }\text{C}$
Temperature of mixture $\left(\theta \right)={32}^{\circ }\text{C}$
Also, the masses of both the liquids are same i.e $m_1=m_2$
From principle of calorimetry, we know that​
Heat gained by liquid 1 = Heat lost by liquid 2
$m_1{s}_1(\theta -{\theta }_1)=m_2{s}_2({\theta }_2-\theta )$
Using the data given in the question, we get
$s_1\times (32-20)=s_2\times (40-32)$
$\Rightarrow s_1\times 12=s_2\times 8$
$\Rightarrow \frac{s_1}{s_2}=\frac{8}{12}=\frac{2}{3}$
Heat capacity of a substance $\left(c\right)=$ mass $\times$ specific heat capacity
i.e $c=ms$
$\therefore \frac{c_1}{c_2}=\frac{s_1}{s_2}=\frac{2}{3}$
Thus, option (d) is the correct answer.