400 g of water at ${100}^{0}C$ is mixed with 200 g of water at ${40}^{0}C$ then, the final temperature of the mixture is

$$\begin{gathered} m_1=400g, t_1={100}^{0}C,m_2=200g \\ {t}_2={40}^{0}C,T=?(final temp) \\ Q lost by hot body =Q gained by cold body \\ {m}_1{s}_w(t_1-T)=m_2{s}_w(T-t_2) \\ \Rightarrow 400\times 1\times (100-T)=200\times 1\times (T-40) \\ \Rightarrow 2(100-T)=T-40\Rightarrow 3T=240 \\ \Rightarrow T={80}^{0}C \\ \therefore T={80}^{0}C \end{gathered}$$