Equilibrium Temperature

Question

There are two container. First container has water at $100°C$ with $200g$, second container has water at $200°C$ with $100g$. How do i calcuate equilibarium temperature? (They are mixed)

Answer 1

The answer is quite simple. You should do a weighted average:$$T=\frac{2}{3}100+\frac{1}{3}200=133.33°C$$ In other words, you have $200 g$ of water at $100°C$ and $100 g$ at $200°C$. So $$T=\frac{200*100+100*200}{300}$$

Answer 2

The exchanged heat quantities are $$ Q_1=m_1c(T-T_1)\\ Q_2=m_2c(T-T_2) $$ And given that $Q_1+Q_2=0$ this implies $$ T=\frac{m_1T_1+m_2T_2}{m_1+m_2} $$