Question

A metal ball of mass 200 g falls from a height of 5 m and hits a slab kept on the ground. If all the kinetic energy of the ball was imparted to the slab as heat energy, what is the rise in temperature of the slab? (Mass of slab : 5 kg, Specific heat capacity of slab = $2J/k{g}^{\circ }C,g=10m/s^2$)

• A. 26°C; 1.25 × 10^-5/°C
• B. 20°C; 1.55 × 10^-5/°C
• C. 25°C; 1.85 × 10^-5/°C
• D. 22°C 1.75/°C

Answer 1

The correct option is A

26°C; 1.25 × 10^-5/°C

When the ball is falling, its gravitational potential energy is converted to kinetic energy.

When it reaches the ground, it would have gained kinetic energy equal to mgh.

So, KE of ball when it hits the slab = decrease in PE = mgh = $0.2\times 5\times 10=10J$

Now, this KE is converted to heat energy,

So , $10=M{C}_c\Delta T$

$\Delta T=\frac{10}{5\times 2}$

$\Delta T=1^{\circ }C$