A heat flux of 4000 J/s is to be passed through a copper rod of length 10 cm and area of cross section `100cm^2`. The thermal conductivity of copper is `400W//m//^(@)C` The two ends of this rod must be kept at a temperature difference of

Options

A. `1^@C`
B. `10^@C`
C. `100^@C`
D. `1000^@C`

Explanation

To solve the problem, we will use the formula for heat transfer through conduction, which is given by Fourier's law:

\[ \frac{dQ}{dt} = -k \cdot A \cdot \frac{\Delta T}{x} \]

Where:
- \(\frac{dQ}{dt}\) is the heat flux (in watts or joules per second),
- \(k\) is the thermal conductivity (in watts per meter per degree Celsius),
- \(A\) is the area of cross-section (in square meters),
- \(\Delta T\) is the temperature difference (in degrees Celsius),
- \(x\) is the length of the rod (in meters).

### Step 1: Identify the given values
- Heat flux, \(\frac{dQ}{dt} = 4000 \, \text{J/s}\)
- Length of the rod, \(x = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m}\)
- Area of cross-section, \(A = 100 \, \text{cm}^2 = 100 \times 10^{-4} \, \text{m}^2 = 0.01 \, \text{m}^2\)
- Thermal conductivity of copper, \(k = 400 \, \text{W/(m} \cdot \text{°C)}\)

### Step 2: Rearrange the formula to solve for \(\Delta T\)
We can rearrange the equation to isolate \(\Delta T\):

\[ \Delta T = \frac{dQ/dt \cdot x}{k \cdot A} \]

### Step 3: Substitute the values into the equation
Now, substitute the known values into the rearranged equation:

\[ \Delta T = \frac{4000 \, \text{J/s} \cdot 0.1 \, \text{m}}{400 \, \text{W/(m} \cdot \text{°C)} \cdot 0.01 \, \text{m}^2} \]

### Step 4: Calculate \(\Delta T\)
Now, perform the calculations:

1. Calculate the numerator:
\[ 4000 \, \text{J/s} \cdot 0.1 \, \text{m} = 400 \, \text{J} \]

2. Calculate the denominator:
\[ 400 \, \text{W/(m} \cdot \text{°C)} \cdot 0.01 \, \text{m}^2 = 4 \, \text{W/°C} \]

3. Now divide the numerator by the denominator:
\[ \Delta T = \frac{400 \, \text{J}}{4 \, \text{W/°C}} = 100 \, \text{°C} \]

### Final Answer
The temperature difference between the two ends of the rod must be \(100 \, \text{°C}\).

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