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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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