Question

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

Answer 1

Given:

The mass of the hot water, $M=20g$

The temperature of the hot water = $80°C$

The mass of the cold water, $m=60g$

The rise in temperature of the cold water $=20°C$

Finding the final temperature of hot water:

We know that the specific heat capacity of water, $c=1Cal{g}^{-1}°C^{-1}$

Let the final temperature of the hot water $=T_{hf}$

From the principle of calorimetry, the heat lost by the hot body = the heat gained by the cold body.

Hence, $Mc(80-T_{hf})=mc\left(20\right)$

$$\begin{gathered} 20\times 1\times (80-T_{hf})=60\times 1\times \left(20\right) \\ \Rightarrow 1600-20T_{hf}=1200 \\ \Rightarrow 400=20T_{hf} \\ \Rightarrow 20=T_{hf} \end{gathered}$$

Hence, the final temperature of the mixture is $20°C$

Finding the initial temperature of cold water:

Let the initial temperature of the cold water $=T_{ci}$

Now, again by substituting the values in the above expression,

$Mc(80-20)=mc(20-T_{ci})$

$$\begin{gathered} 20\times 1\times (80-20)=60\times 1\times (20-T_{ci}) \\ \Rightarrow 1200=1200-60T_{ci} \\ \Rightarrow 0=-60T_{ci} \\ \Rightarrow 0=T_{ci} \end{gathered}$$

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