A uniform copper rod 50 cm long is insulated on the sides, and has its ends exposedto 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 calculating the temperature gradient in a uniform copper rod with its ends exposed to ice and steam, 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 in contact with ice (0°C) and the other end in contact with steam (100°C). There are also 1 mm thick layers of water at each end.

### Step 2: Define the thermal resistances
We will treat the system as a series of thermal resistances. The total thermal resistance \( R \) can be expressed as the sum of the resistances of the water layers and the copper rod.

1. **Resistance of the water layer at the ice end**:
\[ R_{w1} = \frac{d_1}{k_{w}} = \frac{0.001}{0.436} \]
where \( d_1 \) is the thickness of the water layer (1 mm = 0.001 m) and \( k_{w} \) is the thermal conductivity of water.

2. **Resistance of the copper rod**:
\[ R_{c} = \frac{L}{k_{c}} = \frac{0.5}{436} \]
where \( L \) is the length of the copper rod (50 cm = 0.5 m) and \( k_{c} \) is the thermal conductivity of copper.

3. **Resistance of the water layer at the steam end**:
\[ R_{w2} = \frac{d_2}{k_{w}} = \frac{0.001}{0.436} \]
where \( d_2 \) is the thickness of the water layer at the steam end.

### Step 3: Calculate the total resistance
The total resistance \( R_{total} \) is:
\[ R_{total} = R_{w1} + R_{c} + R_{w2} \]
Substituting the values:
\[ R_{total} = \frac{0.001}{0.436} + \frac{0.5}{436} + \frac{0.001}{0.436} \]
Calculating each term:
- \( R_{w1} = \frac{0.001}{0.436} \approx 0.002295 \, \text{K/W} \)
- \( R_{c} = \frac{0.5}{436} \approx 0.001149 \, \text{K/W} \)
- \( R_{w2} = \frac{0.001}{0.436} \approx 0.002295 \, \text{K/W} \)

Thus:
\[ R_{total} \approx 0.002295 + 0.001149 + 0.002295 \approx 0.005739 \, \text{K/W} \]

### Step 4: Calculate the temperature difference
The total temperature difference \( \Delta T \) across the system is:
\[ \Delta T = T_{steam} - T_{ice} = 100°C - 0°C = 100°C \]

### Step 5: Calculate the heat transfer rate \( Q \)
Using Fourier's law of heat conduction:
\[ Q = \frac{\Delta T}{R_{total}} = \frac{100}{0.005739} \approx 17430.5 \, \text{W} \]

### Step 6: Calculate the temperature gradient
The temperature gradient \( \frac{dT}{dx} \) in the copper rod can be calculated using:
\[ \frac{dT}{dx} = \frac{\Delta T}{L} = \frac{100}{0.5} = 200 \, \text{K/m} \]

### Final Answer
The temperature gradient in the copper rod is:
\[ \frac{dT}{dx} = 200 \, \text{K/m} \]