Final Temperature = \(\frac{m_1 T_1 + m_2 T_2}{m_1 + m_2}\) where m\(_1\) and m\(_2\) are the weights of the water in the first and second containers. T\(_1\) and T\(_2\) are the temperatures of the water in the first and second container. Given Final temperature = 50^\circC; Let m\(_1\) = b, then m\(_2\) = 2b. T\(_1\) = t; T\(_2\) = 30^\circC \(\therefore 50 = \frac{b(t) + 2b(30)}{b + 2b}\) \(50 = \frac{bt + 2b(30)}{3b}\) \(50(3b) = bt + 2b(30) \implies 150b = b(t + 60)\) \(150 = t + 60 \implies t = 150 - 60 = 90^\circC\)