Question

A bullet of mass $50\text{g}$ and specific heat capacity $800{\text{J kg}}^{-1}{\text{K}}^{-1}$ is initially at a temperature ${20}^{\circ }\text{C}$. It is fired vertically upwards with a speed of $200\text{m/s}$ and on returning to the starting point strikes a lump of ice at $0^{\circ }\text{C}$ and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice $=3.36\times {10}^5{\text{J kg}}^{-1}$.

• A. Energy of bullet used in melting is $1800\text{J}$
• B. The mass of ice melted $=5\text{g}$
• C. The mass of ice melted is slightly greater than $5\text{g}$
• D. The mass of ice melted is less than $5\text{g}$

Answer 1

Answer: (A), (C)

The mass of ice melted is slightly greater than $5\text{g}$

Speed of bullet will be same when it returns to the starting point [neglecting air drag forces]
K.E. of bullet at starting point is
$KE=\frac{1}{2}m{v}^2=\frac{1}{2}\times 50\times {10}^{-3}\times (200{)}^2$
$\Rightarrow KE=1000\text{J}$
This entire K.E. is used in melting ice.

Heat lost by bullet [as the temperature of bullet falls from ${20}^{\circ }\text{C}$ to $0^{\circ }\text{C}$] $\left(Q\right)=ms\Delta T$
$Q=50\times {10}^{-3}\times 800(20-0)=800\text{J}$

Total energy available with bullet for melting ice
$Q_{total}=KE+Q=1000+800=1800\text{J}$

Let the mass of ice melted be $m\text{kg}$
$\Rightarrow mL=1800$
$\Rightarrow m\times (3.36\times {10}^5)=1800$
$\Rightarrow m=5.35\text{g}$
Hence, options (a) and (c) are the correct answer.