Question

A liquid X of specific heat capacity $1050JK{g}^{-1}{K}^{-1}$ and at $90°C$ is mixed with a liquid Y of specific heat capacity $2362.5JK{g}^{-1}{K}^{-1}$ and at $20°C$, when the final temperature recorded is $50°C$. Find in what proportion the weights of the liquids are mixed.

Answer 1

Step 1: Organised table of the given data

Substance	Mass	S.H.C	Initial Temperature	Final Temperature=$50°C$
Liquid X	$?$	$1050JK{g}^{-1}{K}^{-1}$	$90°C$	${\theta }_F=90-50=40°C$
Liquid Y	$?$	$2362.5JK{g}^{-1}{K}^{-1}$	$20°C$	${\theta }_R=50-20=30°C$

Step 2: Calculating the heat energy

The heat released by the Liquid X is given as

$mc{\theta }_F=X\times 1050\times 40$

The heat absorbed by the Liquid Y is given as

$mc{\theta }_R=Y\times 2362.5\times 30$

Step3: Calculating the proportions of the liquids

Heat released = Heat absorbed

$$\begin{gathered} X\times 1050\times 40=Y\times 2362.5\times 30 \\ \frac{X}{Y}=\frac{27}{16} \end{gathered}$$

Hence, the proportion of the weights of the liquids are mixed is $27:16$