Question

A solid of mass $25g$(sp. heat capacity $0.8J/g°C$) and at $120°C$ is placed in $100g$of water at $20°C$. Calculate the final temperature of the mixture. Specific heat capacity of water is $4.2J/g^{-1}°C$.

Answer 1

Step 1: Given data

Solid: Mass$\left(m\right)=25g$

specific heat capacity$\left(c\right)=0.8J/g^{-1}°C$

Initial temperature$=120°C$

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

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

Initial temperature$=20°C$

Let the final temperature be$x$.

Step 2: calculating the heat absorbed and given out

The change in temperature of solid$\left({\theta }_F\right)=120°C-x$

The change in temperature of water$\left({\theta }_R\right)=x-20°C$

Now, heat is given out by the solid=heat absorbed by the water

$$\begin{gathered} \Rightarrow mc{\theta }_F=mc{\theta }_R \\ \Rightarrow 25\times 0.8\times (120-x)=100\times 4.2\times (x-20) \\ \Rightarrow 2400-20x=420x-8400 \\ \Rightarrow 440x=10800 \\ \Rightarrow x=\frac{10800}{440} \\ \Rightarrow x=24.54°C \end{gathered}$$

The final temperature of the mixture is $24.54°C$.