Question

A metal block of mass 120 g is heated to a temperature of 100$^\circ$C and placed on a huge block of ice at 0$^\circ$C. The specific heat capacity of the metal is 0.12 cal/$^\circ$C and the latent heat of fusion of ice is 80 cal/g. The mass of the ice melted is:

• A. 120 g
• B. 128 g
• C. 115 g
• D. 18 g

Hint:

The energy required to melt ice is proportional to its mass and the latent heat of fusion. The energy transferred from the metal block is used to melt the ice.

Answer 1

The Correct Option is D

Step 1: The energy lost by the metal block will be used to melt the ice. The energy lost by the metal block is given by: \[ Q_{\text{metal}} = m \cdot c \cdot \Delta T \] where \( m = 120 \, g \), \( c = 0.12 \, cal/g ^\circ C \), and \( \Delta T = 100 ^\circ C - 0 ^\circ C = 100 ^\circ C \). \[ Q_{\text{metal}} = 120 \times 0.12 \times 100 = 1440 \, \text{cal} \] 

 

Step 2: The energy required to melt the ice is: \[ Q_{\text{ice}} = m_{\text{ice}} \cdot L_f \] where \( L_f = 80 \, \text{cal/g} \) is the latent heat of fusion and \( m_{\text{ice}} \) is the mass of the ice melted. 

Step 3: Since all the energy from the metal is used to melt the ice: \[ 1440 = m_{\text{ice}} \times 80 \] \[ m_{\text{ice}} = \frac{1440}{80} = 18 \, \text{g} \]