To solve the problem, we will follow these steps:
### Step 1: Understand the given data
- Length of the rod, \( L = 50 \, \text{cm} = 0.5 \, \text{m} \)
- Area of cross-section, \( A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \)
- Thermal conductivity, \( k = 500 \, \text{W/m°C} \)
- Temperature at one end, \( T_1 = 0°C \)
- Temperature at the other end, \( T_2 \) varies from \( 0°C \) to \( 60°C \) over \( 20 \, \text{min} = 1200 \, \text{s} \)
### Step 2: Determine the average temperature difference
Since the temperature at the other end varies linearly, we can find the average temperature difference over the time interval. The temperature at the other end increases linearly from \( 0°C \) to \( 60°C \).
The average temperature difference \( \Delta T \) can be calculated as:
\[
\Delta T = \frac{T_1 + T_2}{2} = \frac{0 + 60}{2} = 30°C
\]
### Step 3: Use the heat conduction formula
The rate of heat transfer \( \frac{dQ}{dt} \) through the rod is given by Fourier's law of heat conduction:
\[
\frac{dQ}{dt} = \frac{k \cdot A \cdot (T_2 - T_1)}{L}
\]
Substituting the average temperature difference:
\[
\frac{dQ}{dt} = \frac{k \cdot A \cdot \Delta T}{L}
\]
### Step 4: Substitute the values
Now substituting the known values:
\[
\frac{dQ}{dt} = \frac{500 \, \text{W/m°C} \cdot 5 \times 10^{-4} \, \text{m}^2 \cdot 30 \, \text{°C}}{0.5 \, \text{m}}
\]
Calculating this:
\[
\frac{dQ}{dt} = \frac{500 \cdot 5 \times 10^{-4} \cdot 30}{0.5}
\]
\[
= \frac{500 \cdot 5 \cdot 30 \times 10^{-4}}{0.5}
\]
\[
= \frac{7500 \times 10^{-4}}{0.5} = 15000 \times 10^{-4} = 1.5 \, \text{W}
\]
### Step 5: Calculate total heat transmitted in 20 minutes
To find the total heat transmitted \( Q \) over \( 1200 \, \text{s} \):
\[
Q = \frac{dQ}{dt} \cdot t = 1.5 \, \text{W} \cdot 1200 \, \text{s}
\]
\[
Q = 1800 \, \text{J}
\]
### Step 6: Convert to kilojoules
\[
Q = \frac{1800}{1000} = 1.8 \, \text{kJ}
\]
### Final Answer
The total heat transmitted through the rod in 20 minutes is **1.8 kJ**.
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