Question

$20g$of hot water at $80°C$ is poured into $60g$ of cold water when the temperature of cold water rises by $20°C$. Calculate the initial temperature of cold water.

Answer 1

Step 1: Organised table of the given data

Substance	Mass	S.H.C	Initial Temperature	Final Temperature=t
Hot water	$20g$	$4.2J{g}^{-1}°C^{-1}$	$80°C$	${\theta }_F=80-t$
Cold water	$60g$	$4.2J{g}^{-1}°C^{-1}$	$?$	${\theta }_R=20°C$

Step 2: Calculating the heat energy

The heat released by the hot water is given as

$mc{\theta }_F=20\times 4.2\times (80-t)=84(80-t)$

The heat absorbed by the cold water is given as

$mc{\theta }_R=60\times 4.2\times 20=5040$

Step3: Calculating the initial temperature of the cold water

The heat released by hot water= Heat absorbed by cold water

$$\begin{gathered} 84(80-t)=5040 \\ 80-t=60 \\ t=20°C \end{gathered}$$

So, the temperature of the mixture is $20°C$

The initial temperature of the cold water$=20-20=0°C$

Hence, the initial temperature of the cold water is $0°C$