A copper rod of length 60 cm long and 8 mm in radius is taken and its one end is kept in boiling water and the other end in ice at `0^(@) C`. If 72 gm of ice melts in one hour with is the thermal conductivity of copper ? Latent of fusion of ice `= 336 "KJ/Kg"`.

Explanation

To find the thermal conductivity of copper using the given data, we can follow these steps:

### Step 1: Convert the given quantities into standard units
- Length of the copper rod, \( L = 60 \text{ cm} = 0.6 \text{ m} \)
- Radius of the copper rod, \( r = 8 \text{ mm} = 0.008 \text{ m} \)
- Mass of ice melted, \( m = 72 \text{ g} = 0.072 \text{ kg} \)
- Latent heat of fusion of ice, \( L_f = 336 \text{ kJ/kg} = 336 \times 10^3 \text{ J/kg} \)
- Time, \( t = 1 \text{ hour} = 3600 \text{ seconds} \)
- Temperature difference, \( \Delta T = 100 \text{ °C} - 0 \text{ °C} = 100 \text{ °C} \)

### Step 2: Calculate the heat absorbed by the ice
Using the formula for heat absorbed:
\[ Q = m \cdot L_f \]
Substituting the values:
\[ Q = 0.072 \text{ kg} \times 336 \times 10^3 \text{ J/kg} = 24192 \text{ J} \]

### Step 3: Calculate the cross-sectional area of the rod
The cross-sectional area \( A \) of the rod can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi \times (0.008 \text{ m})^2 = \pi \times 0.000064 \text{ m}^2 \approx 0.000201 \text{ m}^2 \]

### Step 4: Use the formula for heat conduction
The formula for heat conduction is given by:
\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \cdot t \]
Rearranging for thermal conductivity \( k \):
\[ k = \frac{Q \cdot L}{A \cdot \Delta T \cdot t} \]

### Step 5: Substitute the known values into the equation
Substituting the values we calculated:
\[ k = \frac{24192 \text{ J} \cdot 0.6 \text{ m}}{0.000201 \text{ m}^2 \cdot 100 \text{ °C} \cdot 3600 \text{ s}} \]

### Step 6: Calculate the value of \( k \)
Calculating the denominator:
\[ 0.000201 \text{ m}^2 \cdot 100 \text{ °C} \cdot 3600 \text{ s} = 7236 \text{ J} \]
Now substituting back:
\[ k = \frac{24192 \cdot 0.6}{7236} \approx \frac{14515.2}{7236} \approx 2.008 \text{ W/m°C} \]

### Final Answer
The thermal conductivity of copper is approximately \( k \approx 200.54 \text{ W/m°C} \).

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