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

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Solution

$∵$ Heat loss = Heat gain
$\Rightarrow m_{1} S_{1} {△ T}_{1} = m_{2} S_{2} {△ T}_{2} \Rightarrow m_{1} \cdot 1800 \times (80 - 50) = m_{2} \times 1200 \times (50 - 30) \Rightarrow m_{1} \cdot 1800 \times 30 = m_{2} \times 1200 \times 20 \Rightarrow \frac{m_{1}}{m_{2}} = \frac{1200 \times 20}{1800 \times 30} = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9} \Rightarrow \frac{m_{1}}{m_{2}} = \frac{4}{9}$

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A liquid P of specific heat capacity 1800Jkg−1K−1 and at 80oC is mixed with liquid R of specific heat capacity 1200Jkg−1K−1 and at 30oC. After mixing, the final temperature of mixture is 50oC. In what proportion by weight are the liquids mixed?

∵ Heat loss = Heat gain⇒m1S1△T1=m2S2△T2⇒m1⋅1800×(80−50)=m2×1200×(50−30)⇒m1⋅1800×30=m2×1200×20⇒m1m2=1200×201800×30=23×23=49⇒m1m2=49

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