two liquids at temperature 60 degree c and20 degreec respectively have masses in the ratio 3:4 and their specific heats in the ratio 4:5.if the two liquids are mixed,the resultant temperature is?

Dear Student,

Please find below the solution to the asked query:

Let the masses are m₁ & m₂, the specific heats are C₁ & C₂. Let T be the common temperature.

Then the amount of heat given by the first liquid should be equal to the amount of heat taken by the other liquid.


$$\begin{gathered} m_1{S}_1\left({60}^0-\theta \right)=m_2{S}_2\left(\theta -{20}^0\right)\Rightarrow \frac{m_1}{m_2}\frac{S_1}{S_2}\left({60}^0-\theta \right)=\left(\theta -{20}^0\right) \\ \Rightarrow \frac{3}{4}\frac{4}{5}\left({60}^0-\theta \right)=\left(\theta -{20}^0\right)\Rightarrow {180}^0-3\theta =5\theta -{100}^0 \\ \Rightarrow 8\theta ={280}^0\Rightarrow \theta =\frac{{280}^0}{8}\Rightarrow \theta ={35}^0 \end{gathered}$$

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