A body of length `1 m` having cross sectional area `0.75 m^(2)` has heat flow through it at the rate of `6000 "Joule"//sec`. Then find the temperature difference if `K = 200 Jm^(-1) K^(-1)`.

Options

A. `20^(@)C`
B. `40^(@)C`
C. `80^(@)C`
D. `100^(@)C`

Answer

To solve the problem, we will use the formula for the rate of heat flow through a material, which is given by:

\[ \frac{dq}{dt} = \frac{K \cdot A}{L} \cdot \Delta T \]

Where:
- \(\frac{dq}{dt}\) = rate of heat flow (in Joules per second)
- \(K\) = thermal conductivity (in J/(m·K))
- \(A\) = cross-sectional area (in m²)
- \(L\) = length of the body (in m)
- \(\Delta T\) = temperature difference (in K or °C)

### Step 1: Identify the given values
From the problem, we have:
- \(\frac{dq}{dt} = 6000 \, \text{J/s}\)
- \(K = 200 \, \text{J/(m·K)}\)
- \(A = 0.75 \, \text{m}^2\)
- \(L = 1 \, \text{m}\)

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

\[ \Delta T = \frac{\frac{dq}{dt} \cdot L}{K \cdot A} \]

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

\[ \Delta T = \frac{6000 \, \text{J/s} \cdot 1 \, \text{m}}{200 \, \text{J/(m·K)} \cdot 0.75 \, \text{m}^2} \]

### Step 4: Calculate the denominator
First, calculate the denominator:

\[ 200 \cdot 0.75 = 150 \, \text{J/K} \]

### Step 5: Calculate \(\Delta T\)
Now substitute back into the equation:

\[ \Delta T = \frac{6000}{150} = 40 \, \text{K} \]

### Conclusion
The temperature difference \(\Delta T\) is \(40 \, \text{K}\) (or \(40 \, \text{°C}\) since the size of the temperature difference is the same in Kelvin and Celsius).

### Final Answer
\(\Delta T = 40 \, \text{°C}\)

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