Question

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

Answer 1

Step 1: Given data

Liquid P: specific heat capacity$\left(c\right)=1800Jk{g}^{-1}{K}^{-1}$

Initial temperature$=80°C$

Let mass equal to $p$

Liquid R: specific heat capacity$\left(c\right)=1200Jk{g}^{-1}{K}^{-1}$

Initial temperature$=30°C$

Let mass equal to$r$.

Final temperature$=50°C$

Step 2: calculating the heat gained and lost

The final temperature of liquid P$\left({\theta }_F\right)=(80-50)=30°C$

The final temperature of liquid R$\left({\theta }_R\right)=(50-20)=30°C$

Now, the heat lost by the liquid P= heat gained by the liquid R

$$\begin{gathered} \Rightarrow mc{\theta }_F=mc{\theta }_R \\ \Rightarrow p\times 1800\times 30=r\times 1200\times 30 \\ \Rightarrow \frac{p}{r}=\frac{1200\times 30}{1800\times 30} \\ \Rightarrow \frac{p}{r}=\frac{4}{9} \end{gathered}$$

The liquids are mixed in the proportion of $4:9$by weight.