Question

How many grams of a liquid of specific heat $0.2{\text{cal/g}}^{\circ }\text{C}$ at a temperature ${40}^{\circ }\text{C}$ must be mixed with $100\text{g}$ of a liquid of specific heat $0.5{\text{cal/g}}^{\circ }\text{C}$ at a temperature ${20}^{\circ }\text{C}$, so that the final temperature of the mixture becomes ${32}^{\circ }\text{C}$?

• A. $175\text{gm}$
• B. $300\text{gm}$
• C. $295\text{gm}$
• D. $375\text{gm}$

Answer 1

The correct option is D $375\text{gm}$

Given that,
Mass of liquid 2 $\left(m_2\right)=100\text{gm}$
Specific heat of liquid 1 $\left(c_1\right)=0.2{\text{cal/g}}^{\circ }\text{C}$
Specific heat of liquid 2 $\left(c_2\right)=0.5{\text{cal/g}}^{\circ }\text{C}$
Temperature of liquid 1 $\left({\theta }_1\right)={40}^{\circ }\text{C}$
Temperature of liquid 2 $\left({\theta }_2\right)={20}^{\circ }\text{C}$
Let $m_1$ be the required mass of liquid 1.

When both the liquids are mixed, heat transfer takes place between them until they reach thermal equilibrium at $\theta ={32}^{\circ }\text{C}$
Using principle of calorimetry, we can say that.
Heat lost by liquid 1 = Heat gained by liquid 2
i.e $m_1{c}_1({\theta }_1-\theta )=m_2{c}_2(\theta -{\theta }_2)$
Thus, temperature of mixture can be written as,
$\theta =\frac{m_1{c}_1{\theta }_1+m_2{c}_2{\theta }_2}{m_1{c}_1+m_2{c}_2}$
$\Rightarrow 32=\frac{m_1\times 0.2\times 40+100\times 0.5\times 20}{m_1\times 0.2+100\times 0.5}$
$\Rightarrow 6.4m_1+1600=8m_1+1000$
$\Rightarrow 1.6m_1=600$
$\Rightarrow m_1=375\text{g}$
Thus, option (d) is the correct answer.