A well insulated box is packed with ice at 0°C and kept in a room at a temperature of 30°C. The lid of the box is made of a cardboard sheet of thickness 4 mm and area 100 `"cm"^(2)`. If the ice melts at the rate of `0.045` kg per hour, calculate the thermal conductivity of cardboard assuming that heat can enter the box only through the lid. [L.H. of fusion of ice `= 336 "KJ/kg"` ]

Answer

To solve the problem of calculating the thermal conductivity of cardboard given the melting ice in a well-insulated box, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Temperature of ice (θ1) = 0°C
- Room temperature (θ0) = 30°C
- Thickness of cardboard (x) = 4 mm = 4 × 10^(-3) m
- Area of cardboard (A) = 100 cm² = 100 × 10^(-4) m² = 10^(-2) m²
- Rate of melting of ice (m) = 0.045 kg/hour = 0.045 kg/3600 s
- Latent heat of fusion of ice (L) = 336 kJ/kg = 336 × 10^3 J/kg

2. **Calculate the Heat Required to Melt Ice:**
- The heat (Q) required to melt the ice can be calculated using the formula:
\[ Q = m \cdot L \]
- Substituting the values:
\[ Q = 0.045 \, \text{kg} \times 336 \times 10^3 \, \text{J/kg} = 15120 \, \text{J} \]

3. **Use Fourier's Law of Heat Conduction:**
- According to Fourier's law, the heat transfer through the cardboard lid can be expressed as:
\[ Q = k \cdot A \cdot \frac{(θ0 - θ1)}{x} \cdot t \]
- Where:
- \( k \) = thermal conductivity of cardboard
- \( A \) = area of the lid
- \( θ0 - θ1 \) = temperature difference
- \( x \) = thickness of the cardboard
- \( t \) = time in seconds

4. **Substituting Known Values:**
- From the previous calculations:
- \( Q = 15120 \, \text{J} \)
- \( A = 10^{-2} \, \text{m}^2 \)
- \( θ0 - θ1 = 30 - 0 = 30 \, \text{°C} \)
- \( x = 4 \times 10^{-3} \, \text{m} \)
- \( t = 3600 \, \text{s} \)
- Plugging these values into the equation:
\[ 15120 = k \cdot 10^{-2} \cdot \frac{30}{4 \times 10^{-3}} \cdot 3600 \]

5. **Rearranging to Solve for k:**
- Rearranging the equation gives:
\[ k = \frac{15120 \cdot 4 \times 10^{-3}}{10^{-2} \cdot 30 \cdot 3600} \]

6. **Calculating k:**
- Performing the calculations:
\[ k = \frac{15120 \cdot 4 \times 10^{-3}}{10^{-2} \cdot 30 \cdot 3600} = \frac{60480 \times 10^{-3}}{108000} = 0.056 \, \text{W/m·K} \]

### Final Answer:
The thermal conductivity of cardboard is approximately \( k = 0.056 \, \text{W/m·K} \).