Question

$40g$ of hot water is poured into $100g$of cold water, when the temperature of cold water rises by $10°C$. If the temperature of the hot water is $60°C$, calculate the initial temperature of cold water.

Answer 1

Step 1: Given data

Hot water: mass$\left(m\right)=40g$

Initial temperature$=60°C$

Coldwater: mass $\left(m\right)=100g$

initial temperature=?

specific heat capacity $\left(c\right)=4.2J{g}^{-1}°C^{-1}$

Let final temperature be$x$.

Step 2: calculating the final temperature

The change in temperature of hot water$\left({\theta }_F\right)=(60-x)°C$

The change in temperature of cold water$\left({\theta }_R\right)=10°C$

Now, heat is given out by hot water= heat absorbed by the cold water

$$\begin{gathered} \Rightarrow mc{\theta }_F=mc{\theta }_R \\ \Rightarrow 40\times 4.2\times (60-x)=100\times 4.2\times 10 \\ \Rightarrow 2400-40x=1000 \\ \Rightarrow 40x=2400-1000 \\ \Rightarrow 40x=1400 \\ \Rightarrow x=\frac{1400}{40}=35°C \end{gathered}$$

Now, the initial mixture of cold water$=(35-10)°C=25°C$

The initial temperature of cold water is $25°C$.