To solve the problem of determining how much heat passes through the iron plate each second, we will use the formula for heat transfer through a material, which is derived from Fourier's law of heat conduction. The formula is:
[tex]\[ Q = k \cdot A \cdot \frac{\Delta T}{d} \][/tex]
Where:
- [tex]\( Q \)[/tex] is the heat transfer rate (in Watts, [tex]\( W \)[/tex] or Joules per second, [tex]\( J/s \)[/tex].
- [tex]\( k \)[/tex] is the thermal conductivity of the material (in [tex]\( W/m \cdot K \)[/tex]).
- [tex]\( A \)[/tex] is the cross-sectional area through which heat is being transferred (in [tex]\( m^2 \)[/tex]).
- [tex]\( \Delta T \)[/tex] is the temperature difference across the material (in [tex]\( K \)[/tex] or [tex]\( \degree C \)[/tex]).
- [tex]\( d \)[/tex] is the thickness of the material (in [tex]\( m \)[/tex]).
We have the following given data:
- Thickness of the iron plate, [tex]\( d = 2 \, cm \)[/tex]
- Cross-sectional area, [tex]\( A = 5000 \, cm^2 \)[/tex]
- Temperature on one face, [tex]\( T_1 = 150 \, \degree C \)[/tex]
- Temperature on the other face, [tex]\( T_2 = 140 \, \degree C \)[/tex]
- Thermal conductivity of iron, [tex]\( k = 80 \, W/m \cdot K \)[/tex]
First, convert the thickness and the cross-sectional area from centimeters to meters:
1. Thickness:
[tex]\[ d = \frac{2 \, cm}{100} = 0.02 \, m \][/tex]
2. Cross-sectional Area:
[tex]\[ A = \frac{5000 \, cm^2}{10000} = 0.5 \, m^2 \][/tex]
Next, calculate the temperature difference, [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = T_1 - T_2 = 150 \, \degree C - 140 \, \degree C = 10 \, \degree C \][/tex]
Now, plug these values into the formula to calculate the heat transfer rate:
[tex]\[ Q = 80 \, W/m \cdot K \cdot 0.5 \, m^2 \cdot \frac{10 \, K}{0.02 \, m} \][/tex]
Perform the calculation step-by-step:
- First, calculate the temperature gradient (temperature difference per unit thickness):
[tex]\[ \frac{\Delta T}{d} = \frac{10 \, K}{0.02 \, m} = 500 \, K/m \][/tex]
- Then, multiply the thermal conductivity by the cross-sectional area and the temperature gradient:
[tex]\[ Q = 80 \, W/m \cdot K \cdot 0.5 \, m^2 \cdot 500 \, K/m \][/tex]
[tex]\[ Q = 80 \cdot 0.5 \cdot 500 \][/tex]
[tex]\[ Q = 40 \cdot 500 \][/tex]
[tex]\[ Q = 20000 \, W \][/tex]
Since [tex]\( 1 \, W = 1 \, J/s \)[/tex], the heat transfer rate [tex]\( Q \)[/tex] is [tex]\( 20000 \, J/s \)[/tex], which is also [tex]\( 20000 \, W \)[/tex].
To express the rate in kilojoules per second ([tex]\( kJ/s \)[/tex]):
[tex]\[ Q = 20000 \, J/s \times \frac{1 \, kJ}{1000 \, J} = 20.0 \, kJ/s \][/tex]
Therefore, the heat passing through the plate each second is:
[tex]\[
\boxed{20.0 \, kJ/s}
\][/tex]
