A slab consists of two parallel layers of iron and brass 10 cm and 10.9 cm thick and of thermal conductivities `500 "Wm"^(-1) "K"^(-1) and 109 "Wm"^(-1) "K"^(-1)` respectively. The area of opposite faces are `2m^(2)` each and are at temperatures 373 K and 273 K. Find heat conducted per second across the slab and also the temperature of the interface.

Explanation

To solve the problem step by step, we will first find the temperature at the interface between the two layers and then calculate the heat conducted per second across the slab.

### Step 1: Identify the given data
- Thickness of iron layer, \( L_1 = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Thickness of brass layer, \( L_2 = 10.9 \, \text{cm} = 0.109 \, \text{m} \)
- Thermal conductivity of iron, \( k_1 = 500 \, \text{W/m·K} \)
- Thermal conductivity of brass, \( k_2 = 109 \, \text{W/m·K} \)
- Area of the faces, \( A = 2 \, \text{m}^2 \)
- Temperature at one end (iron side), \( T_1 = 373 \, \text{K} \)
- Temperature at the other end (brass side), \( T_2 = 273 \, \text{K} \)

### Step 2: Set up the equation for heat conduction
The heat conducted per second through both materials is equal, so we can write:

\[ \frac{Q}{t} = \frac{k_1 A (T_1 - T)}{L_1} = \frac{k_2 A (T - T_2)}{L_2} \]

Where \( T \) is the temperature at the interface.

### Step 3: Substitute the known values into the equation
Substituting the values into the equation gives:

\[ \frac{500 \times 2 \times (373 - T)}{0.1} = \frac{109 \times 2 \times (T - 273)}{0.109} \]

### Step 4: Simplify the equation
This simplifies to:

\[ 1000 (373 - T) = 1090 (T - 273) \]

### Step 5: Expand and rearrange the equation
Expanding both sides gives:

\[ 373000 - 1000T = 1090T - 297870 \]

Rearranging the equation to isolate \( T \):

\[ 373000 + 297870 = 1000T + 1090T \]

\[ 670870 = 2090T \]

### Step 6: Solve for \( T \)
Now, divide both sides by 2090 to find \( T \):

\[ T = \frac{670870}{2090} \approx 321.33 \, \text{K} \]

### Step 7: Calculate the heat conducted per second
Now, we can calculate the heat conducted using the temperature at the interface:

Using the equation for heat conduction through iron:

\[ \frac{Q}{t} = \frac{k_1 A (T_1 - T)}{L_1} \]

Substituting the known values:

\[ \frac{Q}{t} = \frac{500 \times 2 \times (373 - 321.33)}{0.1} \]

Calculating the difference in temperature:

\[ 373 - 321.33 = 51.67 \]

Now substituting this back into the equation:

\[ \frac{Q}{t} = \frac{500 \times 2 \times 51.67}{0.1} = \frac{500 \times 2 \times 51.67}{0.1} = 516700 \, \text{W} \]

### Final Result
Thus, the heat conducted per second across the slab is approximately \( 516700 \, \text{J/s} \).