Question

An iron block of mass $2kg$ falls from a height of $10m$ after colliding with the ground it loses$25\%$ of energy to the surrounding then find the temperature rise of the block(take specific heat of $Fe=470J/Kg°C$)

Answer 1

Step 1: Given data-

Mass of iron block $=2Kg$

Height =$10m$

Percentage energy loss $=25\%$

Specific heat of iron $c=470J/Kg°C$

Step 2: Calculation energy-

The initial energy possessed will be equal to the potential energy of block of iron.

Total energy $=mgh$ where $g$is acceleration due to gravity.

$=2\times 10\times 10$

$=200J$

Decrease in energy

$25\%of200=\frac{25}{100}\times 200$$=50J$

Step 3: Calculation for temperature rise-

Let $T$ be the rise in temperature

Loss in energy$=mc∆T$

$$\begin{gathered} \Rightarrow 50=2\times 470\times ∆T \\ \Rightarrow ∆T=\frac{50}{2\times 470} \\ \Rightarrow ∆T=0.05°C \end{gathered}$$

Hence$0.05°C$ is the temperature rise of the block.