One end of a metal rod of length 1.0 m and area of cross section `100 cm^(2)` is maintained at . `100^(@)C`. If the other end of the rod is maintained at `0^(@)C` , the quantity of heat transmitted through the rod per minute is (Coefficient of thermal conductivity of material of rod =`100 W//m-K`)

Options

A. `3xx10^(3)J`
B. `6xx10^(3)J`
C. `9xx10^(3)J`
D. `12xx10^(3)J`

Answer

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

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

Where:
- \( Q \) = quantity of heat transmitted (in Joules)
- \( K \) = coefficient of thermal conductivity (in W/m·K)
- \( A \) = 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: Identify the given values
- Length of the rod, \( L = 1.0 \, m \)
- Area of cross-section, \( A = 100 \, cm^2 = 100 \times 10^{-4} \, m^2 = 0.01 \, m^2 \)
- Coefficient of thermal conductivity, \( K = 100 \, W/m·K \)
- Temperature at one end, \( T_1 = 100 \, °C \)
- Temperature at the other end, \( T_2 = 0 \, °C \)

### Step 2: Calculate the temperature difference
\[ \Delta T = T_1 - T_2 = 100 \, °C - 0 \, °C = 100 \, K \]

### Step 3: Substitute the values into the formula
Using the formula for heat transfer per second:
\[ Q/t = \frac{K \cdot A \cdot \Delta T}{L} \]

Substituting the known values:
\[ Q/t = \frac{100 \, W/m·K \cdot 0.01 \, m^2 \cdot 100 \, K}{1.0 \, m} \]

### Step 4: Calculate the heat transmitted per second
\[ Q/t = \frac{100 \cdot 0.01 \cdot 100}{1} = 100 \, W \]

### Step 5: Convert the heat transmitted per second to heat transmitted per minute
Since we need the quantity of heat transmitted per minute:
\[ Q = Q/t \cdot 60 \, s = 100 \, W \cdot 60 \, s = 6000 \, J \]

### Final Answer
The quantity of heat transmitted through the rod per minute is \( 6000 \, J \).

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