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 will follow these steps:

### Step 1: Understand the Problem
We have a piece of ice at -5 °C, and we are given 420 J of heat. The ice has a specific heat capacity of 2100 J/(kg·°C) and a latent heat of fusion of 3.36 × 10^5 J/kg. We need to find the initial mass of the ice (m grams) given that 1 gram of ice has melted after the heat is added.

### Step 2: Calculate the Heat Required to Raise the Temperature of Ice
First, we need to calculate the heat required to raise the temperature of the ice from -5 °C to 0 °C. The formula for heat (Q) is:

\[ Q_1 = m \cdot c \cdot \Delta T \]

Where:
- \( m \) = mass of ice in kg
- \( c \) = specific heat capacity = 2100 J/(kg·°C)
- \( \Delta T \) = change in temperature = 0 - (-5) = 5 °C

### Step 3: Calculate the Heat Required for Melting
Next, we need to calculate the heat required to melt 1 gram of ice at 0 °C. The formula for heat during phase change is:

\[ Q_2 = m_i \cdot L \]

Where:
- \( m_i \) = mass of ice melted = 1 gram = 0.001 kg
- \( L \) = latent heat of fusion = 3.36 × 10^5 J/kg

### Step 4: Set Up the Equation
The total heat provided (420 J) is equal to the sum of the heat required to raise the temperature of the ice and the heat required to melt the ice:

\[ Q = Q_1 + Q_2 \]

Substituting the values we have:

\[ 420 = (m \cdot 2100 \cdot 5) + (0.001 \cdot 3.36 \times 10^5) \]

### Step 5: Simplify and Solve for m
Now we can simplify and solve for \( m \):

1. Calculate \( Q_2 \):
\[ Q_2 = 0.001 \cdot 3.36 \times 10^5 = 336 \, \text{J} \]

2. Substitute \( Q_2 \) back into the equation:
\[ 420 = (m \cdot 2100 \cdot 5) + 336 \]

3. Rearranging gives:
\[ 420 - 336 = m \cdot 2100 \cdot 5 \]
\[ 84 = m \cdot 10500 \]

4. Finally, solve for \( m \):
\[ m = \frac{84}{10500} \]
\[ m = 0.008 \, \text{kg} = 8 \, \text{grams} \]

### Conclusion
The initial mass of the ice is **8 grams**.

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