Q151 / 180 · COMEDK 2024 (Morning Shift) · Physics
A one $\mathrm{kg}$ block of ice at $-1.5^{\circ} \mathrm{C}$ falls from a height of $1.5 \mathrm{~km}$ and is found melting. The amount of ice melted due to fall, if $60 \%$ energy is converted into heat is (Specific heat capacity of ice is $0.5 \mathrm{~cal} \mathrm{~g}^{-1} \mathrm{~C}^{-1}$, Latent heat of fusion of ice $=80 \mathrm{~cal~g}^{-1}$ )
- A17.9 g
- Correct answerB16.9 g
- C10 g
- D1.69 g
Solution
PE lost $= mgh = 1 \times 9.8 \times 1500 = 14700$ J
Heat generated $= 0.6 \times 14700 = 8820$ J $= \dfrac{8820}{4.2} = 2100$ cal
Heat to warm ice from $-1.5°C$ to $0°C$:
$Q_1 = 1000 \times 0.5 \times 1.5 = 750$ cal
Heat available for melting:
$Q_{melt} = 2100 - 750 = 1350$ cal
Mass melted $= \dfrac{Q_{melt}}{L} = \dfrac{1350}{80} \approx 16.9$ g
Heat generated $= 0.6 \times 14700 = 8820$ J $= \dfrac{8820}{4.2} = 2100$ cal
Heat to warm ice from $-1.5°C$ to $0°C$:
$Q_1 = 1000 \times 0.5 \times 1.5 = 750$ cal
Heat available for melting:
$Q_{melt} = 2100 - 750 = 1350$ cal
Mass melted $= \dfrac{Q_{melt}}{L} = \dfrac{1350}{80} \approx 16.9$ g
Hence, the correct answer is option: 'B'.