  Question

A liquid P of specific heat capacity $$1800 \:J\:kg^{-1}K^{-1}$$ and at $$80\:^oC$$ is mixed with a liquid R of specific heat capacity $$1200\: J\:kg^{-1}K^{-1}$$ and at $$30\:^oC$$. After mixing, the final temperature of mixture is $$50\:^oC$$. In what proportion by weight are the liquids mixed?

Solution

For Liquid P:$$Mass = ?$$$$S.H.C = 1800 \:J\:kg^{-1}K^{-1}$$$$Initial \:Temp. =80\:^oC$$$$Final \:Temp. =50\:^oC$$                          $$=\theta _F=(80-50)=30\:^oC$$for Liquid R:$$Mass = ?$$$$S.H.C = 1200 \:J\:kg^{-1}K^{-1}$$$$Initial \:Temp. =30\:^oC$$$$Final \:Temp. =50\:^oC=\theta _F=(50-30)$$                           $$=20\:^oC$$$$Heat \: lost \: by \: liquid \: P=mc\:\theta _F=P\times 1800\times30$$$$Heat \: gained \: by \: liquid \: R=mc\:\theta _R=R\times 1200\times20$$$$Heat \: gained=Heat \: lost$$$$R\times 1200\times20=P\times 1800\times30$$        $$\dfrac{P}{R}=\dfrac{1200\times 20}{1800\times 30}=\dfrac{4}{9}$$        $$P : R=4 : 9$$Physics

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