Question

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

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

Answer 1

The correct option is C ${60}^{\circ }C$

We know that: $Q=ms\Delta \theta$
Given: First case $m_A=100g,{\theta }_A={100}^{\circ }C$
$m_B=50g,{\theta }_B={75}^{\circ }C,{\theta }_{mix}={90}^{\circ }C$
Second case $m_A=100g,{\theta }_A={100}^{\circ }C,m_B=50g,{\theta }_B={50}^{\circ }C$

In first case heat lost by 𝐴 is absorbed by 𝐵
$m_A{s}_A(100-90)=m_B{s}_B(90-75)$
$100\times s_A\times (100-90)=50\times s_B\times (90-75)$
$2s_A=1.5s_B$
$s_A=\frac{3}{4}{s}_B$ .....(i)

We know that: $Q=ms\Delta \theta$
Heat lost by 𝐴 is absorbed by 𝐵
Now,
$100\times s_A\times (100-T)$
$=50\times s_B\times (T-50)$
From equation (i)
$2\times \frac{3}{4}(100-T)=(T-50)$
$300-3T=2T-100$
$T={80}^{\circ }C$

Final Answer: (b)