Question

If 20 gm of water at $50{}^{o}C$ is mixed with 60 gm of water at $10{}^{o}C$. Then calculate the final temperature of the mixture.

Answer 1

$\theta =\frac{m_1{s}_1{\theta }_1+m_2{s}_2{\theta }_2}{m_1{s}_1+m_2{s}_2}(here{s}_1=s_2)$

$\theta =\frac{20\times 50+60\times 10}{20+60}=\frac{1600}{80}=20{}^{o}C$