Hot water at a temperature of t is added to twice that amount of water at a temperature of 30°C. If the resulting temperature of the mixture is 50°C. Calculate t.
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°C;
Let m\(_1\) = b, then m\(_2\) = 2b.
T\(_1\) = t; T\(_2\) = 30°C
\(\therefore 50 = \frac{b(t) + 2b(30)}{b + 2b}\)
\(50 = \frac{bt + 2b(30)}{3b}\)
\(50(3b) = bt + 2b(30) \implies 150b = b(t + 60)\)
\(150 = t + 60 \implies t = 150 - 60 = 90°C\)
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