Question

$0.1m^3$ of water at ${80}^{\circ }$ is mixed with $0.3m^3$ of water at ${60}^{\circ }$. The final temperature of mixture is :

• A. ${80}^{\circ }C$
• B. ${70}^{\circ }C$
• C. ${60}^{\circ }C$
• D. ${75}^{\circ }C$

Answer 1

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

$m_1{s}_1{dt}_1=m_2{s}_2{dt}_2$

$P\left[0.1\right]\times 1\times [80-t]=P\left[0.3\right]\times \left[1\right]\times [t-60]$

$80-t=3t-180$

$4t=260=t=65^{0}c$