A liquid-nitrogen container is made of a `1-cm` thick styrofoam sheet having thermal conductivity `0.025Js^(-1)m^(-1)C^(-1)` . Liquid nitrogen at ` 80K` is kept in it. A total area of `0.80m^(2)` is in contact with the liquid nitrogen. The atmospheric temperature is `300K` . Calculate the rate of heat flow from the atmosphere to the liquid nitrogen.

Explanation

To calculate the rate of heat flow from the atmosphere to the liquid nitrogen in the container, we can use Fourier's law of heat conduction, which is given by the formula:

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

Where:
- \( Q \) is the rate of heat flow (in watts, W)
- \( K \) is the thermal conductivity of the material (in \( \text{J s}^{-1} \text{m}^{-1} \text{C}^{-1} \))
- \( A \) is the area through which heat is being transferred (in \( \text{m}^2 \))
- \( \Delta T \) is the temperature difference across the material (in degrees Celsius or Kelvin)
- \( d \) is the thickness of the material (in meters)

### Step-by-Step Solution:

1. **Identify the given values:**
- Thickness of the styrofoam sheet, \( d = 1 \text{ cm} = 0.01 \text{ m} \)
- Thermal conductivity, \( K = 0.025 \text{ J s}^{-1} \text{m}^{-1} \text{C}^{-1} \)
- Area, \( A = 0.80 \text{ m}^2 \)
- Temperature of liquid nitrogen, \( T_{\text{liquid}} = 80 \text{ K} \)
- Atmospheric temperature, \( T_{\text{atmosphere}} = 300 \text{ K} \)

2. **Calculate the temperature difference (\( \Delta T \)):**
\[ \Delta T = T_{\text{atmosphere}} - T_{\text{liquid}} = 300 \text{ K} - 80 \text{ K} = 220 \text{ K} \]

3. **Substitute the values into the formula:**
\[ Q = \frac{K \cdot A \cdot \Delta T}{d} \]
\[ Q = \frac{0.025 \, \text{J s}^{-1} \text{m}^{-1} \text{C}^{-1} \cdot 0.80 \, \text{m}^2 \cdot 220 \, \text{K}}{0.01 \, \text{m}} \]

4. **Calculate \( Q \):**
\[ Q = \frac{0.025 \cdot 0.80 \cdot 220}{0.01} \]
\[ Q = \frac{4.4}{0.01} = 440 \, \text{W} \]

5. **Conclusion:**
The rate of heat flow from the atmosphere to the liquid nitrogen is \( 440 \, \text{W} \).