A uniform copper rod 50 cm long is insulated on the sides, and has its ends exposed to ice and steam, respectively. If there is a layer of water 1 mm thick at each end, calculate the temperature gradient in the bar. The thermal conductivity of copper is `436 W m^(-1) K^(-1)` and that of water is `0.436 Wm^(-1) K^(-1)`.

Answer

To solve the problem of finding the temperature gradient in the copper rod, we will follow these steps:

### Step 1: Understand the Setup
We have a copper rod that is 50 cm long, insulated on the sides, with one end exposed to ice (0°C) and the other end exposed to steam (100°C). Each end of the rod has a layer of water that is 1 mm thick.

### Step 2: Convert Units
Convert the lengths from centimeters and millimeters to meters:
- Length of the copper rod, \( L_c = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Thickness of water layer, \( L_w = 1 \, \text{mm} = 0.001 \, \text{m} \)

### Step 3: Calculate Thermal Resistances
The thermal resistance \( R \) for each layer can be calculated using the formula:
\[ R = \frac{L}{kA} \]
where:
- \( L \) is the thickness of the material,
- \( k \) is the thermal conductivity,
- \( A \) is the cross-sectional area (which will cancel out later).

#### Resistance of Water Layers
For one layer of water:
\[ R_w = \frac{L_w}{k_w A} = \frac{0.001}{0.436 A} \]
For two layers of water (since there are two ends):
\[ R_{total \, water} = 2R_w = 2 \times \frac{0.001}{0.436 A} = \frac{0.002}{0.436 A} \]

#### Resistance of Copper Rod
For the copper rod:
\[ R_c = \frac{L_c}{k_c A} = \frac{0.5}{436 A} \]

### Step 4: Total Thermal Resistance
The total thermal resistance \( R_{total} \) is the sum of the resistances:
\[ R_{total} = R_{total \, water} + R_c = \frac{0.002}{0.436 A} + \frac{0.5}{436 A} \]
Combine the terms:
\[ R_{total} = \frac{0.002 + 0.5}{0.436 A} = \frac{0.502}{0.436 A} \]

### Step 5: Calculate Heat Current
The heat current \( Q/t \) can be expressed using the temperature difference and total resistance:
\[ \frac{Q}{t} = \frac{T_{hot} - T_{cold}}{R_{total}} = \frac{100 - 0}{R_{total}} = \frac{100}{\frac{0.502}{0.436 A}} = \frac{100 \cdot 0.436 A}{0.502} \]

### Step 6: Find Temperature Drop Across Copper
Let \( T_1 \) be the temperature at the water-copper interface and \( T_2 \) be the temperature at the copper-ice interface. The temperature drop across the copper rod can be calculated as:
\[ \frac{Q}{t} = \frac{T_2 - T_1}{R_c} \]
Substituting for \( R_c \):
\[ \frac{100 \cdot 0.436 A}{0.502} = \frac{T_2 - T_1}{\frac{0.5}{436 A}} \]

### Step 7: Solve for Temperature Difference \( T_2 - T_1 \)
Cross-multiplying gives:
\[ (T_2 - T_1) = \frac{100 \cdot 0.436 A \cdot 0.5}{0.502} \cdot 436 A \]
This simplifies to:
\[ T_2 - T_1 = \frac{100 \cdot 0.5 \cdot 436^2}{0.502} \]

### Step 8: Calculate Temperature Gradient
The temperature gradient \( \frac{dT}{dx} \) across the copper rod is given by:
\[ \frac{dT}{dx} = \frac{T_2 - T_1}{L_c} = \frac{(T_2 - T_1)}{0.5} \]

### Final Calculation
Calculating the values will yield the temperature gradient.