# 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 *m _{1} & m_{2}*, the specific heats are

*C*. Let

_{1}& C_{2}*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.

${m}_{1}{S}_{1}\left({60}^{0}-\theta \right)={m}_{2}{S}_{2}\left(\theta -{20}^{0}\right)\Rightarrow \frac{{m}_{1}}{{m}_{2}}\frac{{S}_{1}}{{S}_{2}}\left({60}^{0}-\theta \right)=\left(\theta -{20}^{0}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{4}\frac{4}{5}\left({60}^{0}-\theta \right)=\left(\theta -{20}^{0}\right)\Rightarrow {180}^{0}-3\theta =5\theta -{100}^{0}\phantom{\rule{0ex}{0ex}}\Rightarrow 8\theta ={280}^{0}\Rightarrow \theta =\frac{{280}^{0}}{8}\Rightarrow \theta ={35}^{0}$

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