A uniform slab of dimension `10cmxx10cmxx1cm` is kept between two heat reservoir at temperatures `10^(@)C` and `90^(@)C`. The larger surface areas touch the reservoirs. The thermal conductivity of the material is `0.80Wm^(-1)C^(-1)` . Find the amount of heat flowing through the slab per minute.

Explanation

To find the amount of heat flowing through the slab per minute, we can use the formula for heat conduction, which is given by Fourier's law:

\[ Q = \frac{k \cdot A \cdot \Delta T}{d} \]

Where:
- \( Q \) = heat transfer per second (in Joules per second or Watts)
- \( k \) = thermal conductivity of the material (in W/m°C)
- \( A \) = area through which heat is conducted (in m²)
- \( \Delta T \) = temperature difference across the slab (in °C)
- \( d \) = thickness of the slab (in meters)

### Step 1: Convert dimensions to meters
The dimensions of the slab are given as 10 cm x 10 cm x 1 cm. We need to convert these dimensions to meters:
- Length = 10 cm = 0.1 m
- Width = 10 cm = 0.1 m
- Thickness = 1 cm = 0.01 m

### Step 2: Calculate the area (A)
The area \( A \) of the larger surface is:
\[ A = \text{Length} \times \text{Width} = 0.1 \, \text{m} \times 0.1 \, \text{m} = 0.01 \, \text{m}^2 \]

### Step 3: Determine the temperature difference (\( \Delta T \))
The temperature difference between the two reservoirs is:
\[ \Delta T = T_{\text{hot}} - T_{\text{cold}} = 90°C - 10°C = 80°C \]

### Step 4: Use the thermal conductivity (k)
The thermal conductivity \( k \) is given as:
\[ k = 0.80 \, \text{W/m°C} \]

### Step 5: Use the thickness (d)
The thickness \( d \) of the slab is:
\[ d = 0.01 \, \text{m} \]

### Step 6: Substitute values into the heat transfer formula
Now we can substitute all the values into the formula:
\[ Q = \frac{0.80 \, \text{W/m°C} \cdot 0.01 \, \text{m}^2 \cdot 80°C}{0.01 \, \text{m}} \]

### Step 7: Calculate \( Q \)
Calculating \( Q \):
\[ Q = \frac{0.80 \cdot 0.01 \cdot 80}{0.01} = 0.80 \cdot 80 = 64 \, \text{J/s} \]

### Step 8: Convert to heat flow per minute
To find the heat flow per minute, we multiply by 60 seconds:
\[ Q_{\text{per minute}} = 64 \, \text{J/s} \times 60 \, \text{s} = 3840 \, \text{J/min} \]

### Final Answer
The amount of heat flowing through the slab per minute is:
\[ \boxed{3840 \, \text{J/min}} \]