The length, a rod of aluminium is 1.0 m and its area of cross-section is `5.0 cm^(2)`. Its one end is kept at `250^(@)C` and the at `50^(@)C`. How much heat will flow in the rod in 5.0 minutes . K for Al =`2.0xx10^(-1) k J s^(-1) m^(-1) .^(@)C^(-1)` and `J=4.18 cal^(-1)`

Answer

To solve the problem, we will use the formula for heat transfer through conduction:

\[ Q = K \cdot A \cdot \frac{(T_1 - T_2)}{L} \cdot t \]

Where:
- \( Q \) = heat transferred (in Joules)
- \( K \) = thermal conductivity (in kJ/s/m/°C)
- \( A \) = area of cross-section (in m²)
- \( T_1 \) = temperature at one end (in °C)
- \( T_2 \) = temperature at the other end (in °C)
- \( L \) = length of the rod (in m)
- \( t \) = time (in seconds)

### Step 1: Convert the area of cross-section from cm² to m²
Given:
- Area \( A = 5.0 \, \text{cm}^2 \)

Convert to m²:
\[ A = 5.0 \, \text{cm}^2 \times \left(\frac{1 \, \text{m}}{100 \, \text{cm}}\right)^2 = 5.0 \times 10^{-4} \, \text{m}^2 \]

### Step 2: Identify the given values
- Length of the rod \( L = 1.0 \, \text{m} \)
- Temperature difference \( T_1 - T_2 = 250 - 50 = 200 \, \text{°C} \)
- Thermal conductivity \( K = 2.0 \times 10^{-1} \, \text{kJ/s/m/°C} = 200 \, \text{J/s/m/°C} \) (since \( 1 \, \text{kJ} = 1000 \, \text{J} \))
- Time \( t = 5.0 \, \text{minutes} = 5 \times 60 = 300 \, \text{s} \)

### Step 3: Substitute the values into the formula
Now, substituting the values into the heat transfer equation:
\[ Q = K \cdot A \cdot \frac{(T_1 - T_2)}{L} \cdot t \]
\[ Q = 200 \, \text{J/s/m/°C} \cdot 5.0 \times 10^{-4} \, \text{m}^2 \cdot \frac{200 \, \text{°C}}{1.0 \, \text{m}} \cdot 300 \, \text{s} \]

### Step 4: Calculate \( Q \)
Calculating step-by-step:
1. Calculate \( K \cdot A \):
\[ 200 \cdot 5.0 \times 10^{-4} = 0.1 \, \text{J/°C} \]

2. Calculate \( \frac{(T_1 - T_2)}{L} \):
\[ \frac{200}{1} = 200 \, \text{°C/m} \]

3. Combine the results:
\[ Q = 0.1 \cdot 200 \cdot 300 \]
\[ Q = 6000 \, \text{J} \]

### Step 5: Convert Joules to calories
Using the conversion \( 1 \, \text{cal} = 4.18 \, \text{J} \):
\[ Q = \frac{6000 \, \text{J}}{4.18 \, \text{J/cal}} \approx 1435.4 \, \text{cal} \]

### Final Answer
The total heat that will flow in the rod in 5.0 minutes is approximately \( 1435.4 \, \text{cal} \).