Question

A one kg block of ice at $ -1.5^\circ C $ falls from a height of 1.5 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 cal g$^{-1}$ C$^{-1}$, Latent heat of fusion of ice = 80 cal g$^{-1}$)

• A. 1.69 g
• B. 10 g
• C. 16.9 g
• D. 17.9 g

Hint:

When calculating the energy converted to heat, use the potential energy of the object falling and the latent heat of fusion to find the mass of ice melted. Don't forget to convert units if necessary (like from cal to J).

Answer 1

The Correct Option is C

The energy converted into heat is due to the potential energy of the ice block as it falls. 
The potential energy \( E_p \) is given by the formula: \[ E_p = mgh \] Where: 
- \( m = 1000 \, \text{g} \) (mass of the ice block), 
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), 
- \( h = 1.5 \, \text{km} = 1500 \, \text{m} \) (height). 
So, \[ E_p = 1000 \times 9.8 \times 1500 = 1.47 \times 10^7 \, \text{J} \] This energy is partially converted to heat. Since 60% of the energy is converted into heat, the heat energy \( Q \) is: \[ Q = 0.6 \times 1.47 \times 10^7 = 8.82 \times 10^6 \, \text{J} \] Now, to convert this energy into heat, we use the specific heat capacity and the latent heat of fusion. 
The amount of ice melted \( m_{\text{melted}} \) is given by the relation: \[ Q = m_{\text{melted}} \times L_f \] Where: - \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion), - \( 1 \, \text{cal} = 4.184 \, \text{J} \). So, \[ m_{\text{melted}} = \frac{Q}{L_f \times 4.184} = \frac{8.82 \times 10^6}{80 \times 4.184} = 16.9 \, \text{g} \] Thus, the amount of ice melted is \( 16.9 \, \text{g} \).