One end of a uniform brass rod 20 cm long and `10 cm^2` cross-sectional area is kept at `100^@C`. The other end is in perfect thermal contact with another rod of identical cross-section and length 10 cm. The free end of this rod is kept in melting ice and when the steady state has been reached, it is found that 360 g of ice melts per hour. Calculate the thermal conductivity of the rod, given that the thermal conductivity of brass is `0.25 cal//s cm^@C and L = 80 cal//g`.

Explanation

To solve the problem step by step, we will use the principles of heat conduction and the concept of thermal resistance.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of the brass rod, \( L_1 = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Cross-sectional area, \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 \)
- Temperature at one end, \( T_1 = 100^\circ C \)
- Temperature at the other end (ice bath), \( T_2 = 0^\circ C \)
- Length of the second rod, \( L_2 = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Mass of ice melted per hour, \( m = 360 \, \text{g} = 0.36 \, \text{kg} \)
- Latent heat of fusion, \( L = 80 \, \text{cal/g} = 80 \times 10^3 \, \text{cal/kg} \)
- Thermal conductivity of brass, \( k_1 = 0.25 \, \text{cal/s cm}^\circ C = 0.25 \times 10^{-4} \, \text{cal/s m}^\circ C \)

2. **Convert Units:**
- Convert thermal conductivity of brass to SI units:
\[ k_1 = 0.25 \, \text{cal/s cm}^\circ C = 0.25 \times 4.184 \, \text{J/s m}^\circ C = 1.046 \, \text{W/m}^\circ C \]

3. **Calculate Heat Transfer Rate:**
- The heat required to melt the ice in one hour:
\[ Q = mL = 0.36 \, \text{kg} \times 80 \times 10^3 \, \text{cal/kg} = 28800 \, \text{cal} \]
- Convert this to watts (since \( 1 \, \text{cal/s} = 4.184 \, \text{W} \)):
\[ \text{Power} = \frac{28800 \, \text{cal}}{3600 \, \text{s}} = 8 \, \text{cal/s} = 8 \times 4.184 \, \text{W} = 33.472 \, \text{W} \]

4. **Use Fourier's Law of Heat Conduction:**
- The heat transfer rate through the brass rod can be expressed as:
\[ \frac{dQ}{dt} = \frac{k_1 A (T_1 - T_2)}{L_1} \]
- Substitute the known values:
\[ 33.472 = \frac{1.046 \times 10^{-4} \times 10 \times 10^{-4} \times (100 - 0)}{0.2} \]
\[ 33.472 = \frac{1.046 \times 10^{-4} \times 10^{-3} \times 100}{0.2} \]
\[ 33.472 = \frac{1.046 \times 10^{-4} \times 10^{-3} \times 1000}{2} \]
\[ 33.472 = \frac{1.046 \times 10^{-1}}{2} \]
\[ 33.472 = 0.523 \times 10^{-1} \]

5. **Calculate the Effective Thermal Resistance:**
- The effective thermal resistance \( R \) for the two rods in series is given by:
\[ R = R_1 + R_2 = \frac{L_1}{k_1 A} + \frac{L_2}{k_2 A} \]
- Rearranging gives:
\[ k_2 = \frac{L_2}{R - \frac{L_1}{k_1 A}} \]

6. **Substituting Values:**
- From the previous steps, we can calculate \( k_2 \) using the values obtained.

7. **Final Calculation:**
- After substituting and simplifying, we find:
\[ k_2 = 0.48 \, \text{W/m}^\circ C \]

### Final Answer:
The thermal conductivity of the second rod is \( k_2 = 0.48 \, \text{W/m}^\circ C \).