One end of a steel rod `(K=46Js^(-1)m^(-1)C^(-1))` of length `1.0m` is kept in ice at `0^(@)C` and the other end is kept in boiling water at `100^(@)C` . The area of cross section of the rod is `0.04cm^(2)` . Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice `=3.36xx10^(5)Jkg^(-1)` .

Explanation

To solve the problem of finding the mass of ice melting per second when one end of a steel rod is kept in ice at 0°C and the other end in boiling water at 100°C, we can follow these steps:

### Step 1: Understand the Heat Transfer Equation
The heat conducted through the rod can be calculated using the formula:
\[ Q = \frac{K \cdot A \cdot (T_1 - T_2)}{L} \]
where:
- \( Q \) = heat transfer per second (in Joules per second or Watts)
- \( K \) = thermal conductivity of the material (in \( \text{Js}^{-1}\text{m}^{-1}\text{C}^{-1} \))
- \( A \) = cross-sectional area of the rod (in \( \text{m}^2 \))
- \( T_1 \) = temperature at one end (in °C)
- \( T_2 \) = temperature at the other end (in °C)
- \( L \) = length of the rod (in meters)

### Step 2: Convert Units
Convert the area from cm² to m²:
\[ A = 0.04 \, \text{cm}^2 = 0.04 \times 10^{-4} \, \text{m}^2 = 4 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Substitute Values into the Heat Transfer Equation
Substituting the known values into the equation:
- \( K = 46 \, \text{Js}^{-1}\text{m}^{-1}\text{C}^{-1} \)
- \( A = 4 \times 10^{-6} \, \text{m}^2 \)
- \( T_1 = 100 \, \text{°C} \)
- \( T_2 = 0 \, \text{°C} \)
- \( L = 1 \, \text{m} \)

The heat transfer per second \( Q \) is:
\[ Q = \frac{46 \cdot (4 \times 10^{-6}) \cdot (100 - 0)}{1} \]
\[ Q = 46 \cdot (4 \times 10^{-6}) \cdot 100 \]
\[ Q = 46 \cdot 4 \times 10^{-4} = 184 \times 10^{-4} = 1.84 \times 10^{-2} \, \text{J/s} \]

### Step 4: Relate Heat Transfer to Ice Melting
The heat required to melt ice can be expressed as:
\[ Q = \dot{m} \cdot L_f \]
where:
- \( \dot{m} \) = mass of ice melting per second (in kg/s)
- \( L_f \) = latent heat of fusion of ice (in J/kg)

Given \( L_f = 3.36 \times 10^5 \, \text{J/kg} \), we can rearrange the equation to find \( \dot{m} \):
\[ \dot{m} = \frac{Q}{L_f} \]

### Step 5: Substitute \( Q \) and Calculate \( \dot{m} \)
Substituting the values:
\[ \dot{m} = \frac{1.84 \times 10^{-2}}{3.36 \times 10^5} \]
\[ \dot{m} = \frac{1.84 \times 10^{-2}}{3.36 \times 10^5} \approx 5.47 \times 10^{-8} \, \text{kg/s} \]

### Step 6: Convert to Grams
To convert the mass from kg/s to grams/s:
\[ \dot{m} = 5.47 \times 10^{-8} \, \text{kg/s} \times 1000 \, \text{g/kg} = 5.47 \times 10^{-5} \, \text{g/s} \]

### Final Answer
The mass of ice melting per second is approximately:
\[ \dot{m} \approx 5.47 \times 10^{-5} \, \text{g/s} \]

---