One end of a 0.25 m long metal bar is in steam and the other is in contact with ice . If 12 g of ice melts per minute, what is the thermal conductivity of the metal? Given cross-section of the bar `=5xx10^(-4) m^(2)` and latent heat of ice is `80 cal g^(-1)`

Options

A. `20 cal s^(-1) m^(-1).^(@)C^(-1)`
B. `10 cal s^(-1) m^(-1) .^(@)C^(-1)`
C. `40 cal s^(-1) m^(-1) .^(@)C^(-1)`
D. `80 cal s^(-1) m^(-1) .^(@)C^(-1)`

Explanation

To find the thermal conductivity of the metal bar, we can use the formula for heat conduction, which is given by Fourier's law:

\[ Q = \frac{k \cdot A \cdot (T_1 - T_2) \cdot t}{L} \]

Where:
- \( Q \) is the heat transferred (in calories),
- \( k \) is the thermal conductivity (in calories per second per meter per degree Celsius),
- \( A \) is the cross-sectional area (in square meters),
- \( T_1 - T_2 \) is the temperature difference (in degrees Celsius),
- \( t \) is the time (in seconds),
- \( L \) is the length of the bar (in meters).

### Step 1: Identify the given values
- Length of the metal bar, \( L = 0.25 \, \text{m} \)
- Cross-sectional area, \( A = 5 \times 10^{-4} \, \text{m}^2 \)
- Temperature at one end (steam), \( T_1 = 100 \, \text{°C} \)
- Temperature at the other end (ice), \( T_2 = 0 \, \text{°C} \)
- Mass of ice melted per minute, \( m = 12 \, \text{g} \)
- Latent heat of ice, \( L = 80 \, \text{cal/g} \)
- Time taken to melt the ice, \( t = 1 \, \text{min} = 60 \, \text{s} \)

### Step 2: Calculate the heat required to melt the ice
Using the formula for heat:

\[ Q = m \cdot L \]

Substituting the values:

\[ Q = 12 \, \text{g} \cdot 80 \, \text{cal/g} = 960 \, \text{cal} \]

### Step 3: Calculate the temperature difference
The temperature difference \( T_1 - T_2 \) is:

\[ T_1 - T_2 = 100 \, \text{°C} - 0 \, \text{°C} = 100 \, \text{°C} \]

### Step 4: Substitute the values into Fourier's law
Now substituting all known values into the Fourier's law equation:

\[ 960 = \frac{k \cdot (5 \times 10^{-4}) \cdot (100) \cdot (60)}{0.25} \]

### Step 5: Rearranging the equation to solve for \( k \)
Rearranging gives:

\[ k = \frac{960 \cdot 0.25}{(5 \times 10^{-4}) \cdot (100) \cdot (60)} \]

### Step 6: Calculate \( k \)
Calculating the right-hand side:

\[ k = \frac{960 \cdot 0.25}{(5 \times 10^{-4}) \cdot 100 \cdot 60} \]

Calculating the denominator:

\[ (5 \times 10^{-4}) \cdot 100 \cdot 60 = 3 \, \text{cal} \]

Now substituting back:

\[ k = \frac{240}{3} = 80 \, \text{cal/s/m/°C} \]

### Final Answer
The thermal conductivity of the metal bar is:

\[ k = 80 \, \text{cal/s/m/°C} \]