An iron plate `10^(-5) m^(2)` area, `4 xx 10^(-3)` m thick has its opposite faces maintained at 373 K and 223 K respectively. How much heat flows through the plate per second ? Thermal conductivity of iron `= 80 "Wm"^(-1) "K"^(-1)`.

Explanation

To solve the problem of how much heat flows through the iron plate per second, we can use the formula for heat transfer through conduction, which is given by:

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

Where:
- \( Q \) = heat transferred (in Joules)
- \( k \) = thermal conductivity (in W/m·K)
- \( A \) = area (in m²)
- \( T_1 \) = temperature of one face (in K)
- \( T_2 \) = temperature of the opposite face (in K)
- \( t \) = time (in seconds)
- \( x \) = thickness of the plate (in m)

### Step 1: Write down the given values
- Area, \( A = 10^{-5} \, m^2 \)
- Thickness, \( x = 4 \times 10^{-3} \, m \)
- Temperature of one face, \( T_1 = 373 \, K \)
- Temperature of the opposite face, \( T_2 = 223 \, K \)
- Thermal conductivity of iron, \( k = 80 \, W/m·K \)
- Time, \( t = 1 \, s \)

### Step 2: Calculate the temperature difference
Calculate the temperature difference \( (T_1 - T_2) \):
\[ T_1 - T_2 = 373 \, K - 223 \, K = 150 \, K \]

### Step 3: Substitute the values into the formula
Now substitute the values into the heat transfer formula:
\[ Q = \frac{80 \, W/m·K \cdot 10^{-5} \, m^2 \cdot 150 \, K \cdot 1 \, s}{4 \times 10^{-3} \, m} \]

### Step 4: Simplify the equation
Now simplify the equation step by step:
1. Calculate the numerator:
\[ 80 \cdot 10^{-5} \cdot 150 \cdot 1 = 1200 \cdot 10^{-5} = 1.2 \times 10^{-3} \, W·s \]
2. Divide by the thickness:
\[ Q = \frac{1.2 \times 10^{-3}}{4 \times 10^{-3}} = 0.3 \, J/s \]

### Step 5: Final answer
Thus, the amount of heat flowing through the plate per second is:
\[ Q = 30 \, J \]