Question

When $100\text{gms}$ of a liquid $A$, at $100{}^{\circ }\text{C}$ is added to $50\text{gms}$ of a liquid $B$ at temperature $75{}^{\circ }\text{C}$, the temperature of the mixture becomes $90{}^{\circ }\text{C}$. The temperature of the mixture, if $100\text{gms}$ of liquid $A$ at $100{}^{\circ }\text{C}$ is added to $50\text{gms}$ of liquid $B$ at $50{}^{\circ }\text{C}$ will be,

• A. $85{}^{\circ }\text{C}$
• B. $80{}^{\circ }\text{C}$
• C. $70{}^{\circ }\text{C}$
• D. $60{}^{\circ }\text{C}$

Answer 1

The correct option is B $80{}^{\circ }\text{C}$

$\text{Heat lost by liquid A = Heat gained by liquid B}$

$\text{So,}{m}_A{S}_A\Delta {\theta }_A=m_B{S}_B\Delta {\theta }_B$

$\Rightarrow 100\times S_A\times (100-90)=50\times S_B\times (90-75)$

$20S_A=15S_B\Rightarrow S_A=\frac{3}{4}{S}_B$

$\text{Similarly,}100\times S_A\times (100-\theta )=50\times S_B\times (\theta -50)$

$2\times \left(\frac{3}{4}\right)S_B\times (100-\theta )=S_B(\theta -50)$

$300-3\theta =2\theta -100$

$\Rightarrow \theta ={80}^{\circ }\text{C}$

Hence, $\left(C\right)$ is the correct answer.