Work done in converting 1 g of ice at `-10^@C` into steam at `100^@C` is

Options

A. 3045 J
B. 6056 J
C. 721 J
D. 616 J

Explanation

To find the work done in converting 1 g of ice at -10°C into steam at 100°C, we need to calculate the total heat required for each step of the phase change and temperature increase. Here's a step-by-step breakdown of the process:

### Step 1: Heating Ice from -10°C to 0°C
- **Formula**: \( q_1 = m \cdot s \cdot \Delta T \)
- **Where**:
- \( m = 1 \, \text{g} \) (mass of ice)
- \( s = \frac{1}{2} \, \text{cal/g°C} \) (specific heat of ice)
- \( \Delta T = 10 \, \text{°C} \) (temperature change from -10°C to 0°C)

**Calculation**:
\[ q_1 = 1 \, \text{g} \cdot \frac{1}{2} \, \text{cal/g°C} \cdot 10 \, \text{°C} = 5 \, \text{cal} \]

### Step 2: Melting Ice at 0°C to Water at 0°C
- **Formula**: \( q_2 = m \cdot L_f \)
- **Where**:
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)

**Calculation**:
\[ q_2 = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 3: Heating Water from 0°C to 100°C
- **Formula**: \( q_3 = m \cdot s \cdot \Delta T \)
- **Where**:
- \( s = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100 \, \text{°C} \) (temperature change from 0°C to 100°C)

**Calculation**:
\[ q_3 = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 100 \, \text{cal} \]

### Step 4: Converting Water at 100°C to Steam at 100°C
- **Formula**: \( q_4 = m \cdot L_v \)
- **Where**:
- \( L_v = 540 \, \text{cal/g} \) (latent heat of vaporization)

**Calculation**:
\[ q_4 = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal} \]

### Step 5: Total Heat Required
Now, we sum up all the heat quantities calculated:
\[ Q_{total} = q_1 + q_2 + q_3 + q_4 \]
\[ Q_{total} = 5 \, \text{cal} + 80 \, \text{cal} + 100 \, \text{cal} + 540 \, \text{cal} = 725 \, \text{cal} \]

### Step 6: Convert Calories to Joules
To convert calories to joules, we use the conversion factor \( 1 \, \text{cal} = 4.2 \, \text{J} \):
\[ Q_{total} = 725 \, \text{cal} \times 4.2 \, \text{J/cal} = 3045 \, \text{J} \]

### Final Answer
The work done in converting 1 g of ice at -10°C into steam at 100°C is **3045 J**.

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