A piece of ice (heat capacity `=2100 J kg^(-1) .^(@)C^(-1)` and latent heat `=3.36xx10^(5) J kg^(-1)`) of mass m grams is at `-5 .^(@)C` at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice . Water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m in gram is

Explanation

To solve the problem, we need to calculate the mass \( m \) of the ice initially present using the heat provided and the properties of the ice. Here’s a step-by-step solution:

### Step 1: Understand the heat transfer involved
When heat is added to the ice, it first raises the temperature of the ice from \(-5^\circ C\) to \(0^\circ C\), and then it melts some of the ice at \(0^\circ C\). The total heat provided is used for both processes.

### Step 2: Calculate the heat required to raise the temperature of the ice
The heat required to raise the temperature of \( m \) kg of ice from \(-5^\circ C\) to \(0^\circ C\) can be calculated using the formula:
\[ Q_1 = m \cdot C \cdot \Delta T \]
where:
- \( C = 2100 \, J \, kg^{-1} \, °C^{-1} \) (specific heat capacity of ice)
- \( \Delta T = 0 - (-5) = 5 \, °C \)

Thus,
\[ Q_1 = m \cdot 2100 \cdot 5 \]

### Step 3: Calculate the heat required to melt the ice
The heat required to melt \( 1 \, g \) (or \( 0.001 \, kg \)) of ice at \(0^\circ C\) is given by:
\[ Q_2 = m_{ice} \cdot L \]
where:
- \( L = 3.36 \times 10^5 \, J \, kg^{-1} \) (latent heat of fusion)
- \( m_{ice} = 0.001 \, kg \)

Thus,
\[ Q_2 = 0.001 \cdot 3.36 \times 10^5 \]

### Step 4: Set up the equation for total heat
The total heat provided is \( 420 \, J \), which is used for both raising the temperature and melting the ice:
\[ Q_1 + Q_2 = 420 \]

Substituting the expressions for \( Q_1 \) and \( Q_2 \):
\[ m \cdot 2100 \cdot 5 + 0.001 \cdot 3.36 \times 10^5 = 420 \]

### Step 5: Solve for \( m \)
Substituting \( Q_2 \):
\[ m \cdot 10500 + 336 = 420 \]
\[ m \cdot 10500 = 420 - 336 \]
\[ m \cdot 10500 = 84 \]
\[ m = \frac{84}{10500} \]
\[ m = 0.008 \, kg = 8 \, g \]

Thus, the value of \( m \) is \( 8 \, g \).

### Final Answer:
The mass \( m \) of the ice initially present is **8 grams**.