One end of a steel rod of length 1 m and area of cross section `4 xx 10^(-6)m^(2)` is put in boiling water and the other end is kept in an ice bath at `0^(@)C.` If thermal conductivity of steel is `46 W//m^(@)C,` find the amount of ice melting per second if heat flow only by conduction. Given that the latent heat of fusion of ice is `3.36 xx 10^(5)J//kg.`

Answer

To solve the problem, we will use the concept of heat transfer by conduction. The formula for heat transfer through conduction is given by:

\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \cdot t \]

Where:
- \( Q \) = Heat transferred (in Joules)
- \( K \) = Thermal conductivity of the material (in W/m°C)
- \( A \) = Area of cross-section (in m²)
- \( \Delta T \) = Temperature difference (in °C)
- \( L \) = Length of the rod (in meters)
- \( t \) = Time (in seconds)

### Step 1: Identify the given values
- Length of the steel rod, \( L = 1 \, m \)
- Area of cross-section, \( A = 4 \times 10^{-6} \, m^2 \)
- Thermal conductivity of steel, \( K = 46 \, W/m°C \)
- Temperature at one end (boiling water), \( T_1 = 100°C \)
- Temperature at the other end (ice bath), \( T_2 = 0°C \)
- Latent heat of fusion of ice, \( L_f = 3.36 \times 10^5 \, J/kg \)

### Step 2: Calculate the temperature difference
\[ \Delta T = T_1 - T_2 = 100°C - 0°C = 100°C \]

### Step 3: Substitute values into the heat transfer formula
We need to find the heat transfer per second, so we set \( t = 1 \, s \):
\[ Q = \frac{K \cdot A \cdot \Delta T}{L} \cdot 1 \]
Substituting the known values:
\[ Q = \frac{46 \, W/m°C \cdot 4 \times 10^{-6} \, m^2 \cdot 100°C}{1 \, m} \]

### Step 4: Calculate \( Q \)
\[ Q = 46 \cdot 4 \times 10^{-6} \cdot 100 \]
\[ Q = 18400 \times 10^{-6} \, J \]
\[ Q = 0.0184 \, J/s \]

### Step 5: Calculate the mass of ice melted per second
Using the latent heat of fusion:
\[ \text{mass melted per second} = \frac{Q}{L_f} \]
Substituting the values:
\[ \text{mass melted per second} = \frac{0.0184 \, J/s}{3.36 \times 10^5 \, J/kg} \]
\[ \text{mass melted per second} = 5.47 \times 10^{-5} \, kg/s \]

### Step 6: Convert to grams
\[ \text{mass melted per second} = 5.47 \times 10^{-5} \, kg/s \times 1000 \, g/kg = 5.47 \times 10^{-2} \, g/s \]

### Final Answer
The amount of ice melting per second is approximately:
\[ 5.47 \times 10^{-2} \, g/s \]