A cylindrical copper rod of length 0.60m and cross-sectional area 6.0 `cm^(2)` is insulated along its side. The ends are held at a temperature difference of `100^(@)C` by having one end in a water -ice mixture and the other in a mixture of boiling water and steam. At what reate (a) is energy conducted by the rod and (b) does the ice melt ?

Answer

To solve the problem step by step, we will first calculate the rate of energy conducted by the copper rod and then determine the rate at which the ice melts.

### Given Data:
- Length of the copper rod, \( L = 0.60 \, \text{m} \)
- Cross-sectional area, \( A = 6.0 \, \text{cm}^2 = 6.0 \times 10^{-4} \, \text{m}^2 \)
- Temperature difference, \( \Delta T = 100 \, ^\circ C \)
- Thermal conductivity of copper, \( k = 401 \, \text{W/m} \cdot \text{K} \)

### Part (a): Rate of Energy Conducted by the Rod

1. **Use Fourier's Law of Heat Conduction**:
\[ P = \frac{k \cdot A \cdot (T_H - T_C)}{L} \]
Where:
- \( P \) is the rate of heat transfer (W)
- \( k \) is the thermal conductivity (W/m·K)
- \( A \) is the cross-sectional area (m²)
- \( T_H - T_C \) is the temperature difference (K)
- \( L \) is the length of the rod (m)

2. **Substitute the values into the formula**:
\[ P = \frac{401 \, \text{W/m·K} \cdot 6.0 \times 10^{-4} \, \text{m}^2 \cdot 100 \, \text{K}}{0.60 \, \text{m}} \]

3. **Calculate the numerator**:
\[ 401 \cdot 6.0 \times 10^{-4} \cdot 100 = 24060 \, \text{W·m}^2 \]

4. **Calculate the rate of heat transfer**:
\[ P = \frac{24060}{0.60} = 40100 \, \text{W} \]

5. **Convert to Joules per second**:
\[ P = 40 \, \text{J/s} \]

### Part (b): Rate at which the Ice Melts

1. **Use the formula for the rate of melting ice**:
\[ \frac{dm}{dt} = \frac{P}{L_f} \]
Where:
- \( \frac{dm}{dt} \) is the mass of ice melted per second (g/s)
- \( L_f \) is the latent heat of fusion of ice (approximately \( 333 \, \text{J/g} \))

2. **Substitute the values**:
\[ \frac{dm}{dt} = \frac{40 \, \text{J/s}}{333 \, \text{J/g}} \]

3. **Calculate the rate of melting**:
\[ \frac{dm}{dt} \approx 0.12012 \, \text{g/s} \approx 0.12 \, \text{g/s} \]

### Final Answers:
- (a) The rate of energy conducted by the rod is **40 J/s**.
- (b) The rate at which the ice melts is **0.12 g/s**.