Three identical cubical boxes each of side 100cm are filled with ice at 0⁰C. The are closed on all sides and stacked one over the other inside a large hall with air temperature of 30⁰C. All the walls are of same thickness [0.3cm]. IF 300gm of ice melts each second in the middle box, the thermal conductivity of material of the box is [assume latent heat of ice=3.33 X 10⁵J/Kg] ?

To find the thermal conductivity of the material of the boxes, we can use the concept of heat transfer through conduction. The heat required to melt the ice in the middle box can be calculated using the formula for latent heat, and then we can relate this to the heat transfer through the walls of the box.

Understanding the Problem

We have three identical cubical boxes, each with a side length of 100 cm (or 1 meter). The boxes are filled with ice at 0 °C and are stacked in a hall where the air temperature is 30 °C. The walls of the boxes are 0.3 cm thick. We know that 300 grams of ice melts each second in the middle box.

Calculating the Heat Required to Melt Ice

The latent heat of fusion for ice is given as 3.33 x 10⁵ J/kg. To find the heat required to melt 300 grams of ice, we first convert grams to kilograms:

• 300 g = 0.3 kg

Now, we can calculate the heat (Q) required to melt this amount of ice:

Q = mass × latent heat

Substituting the values:

Q = 0.3 kg × 3.33 x 10⁵ J/kg = 99900 J

This means that 99900 Joules of heat is absorbed by the ice every second to melt it.

Heat Transfer Through the Walls of the Box

The heat transfer through the walls of the box can be described by Fourier's law of heat conduction:

Q = k × A × (ΔT) / d

• Q = heat transfer per unit time (J/s or Watts)
• k = thermal conductivity of the material (W/m·K)
• A = surface area of the box (m²)
• ΔT = temperature difference across the wall (K)
• d = thickness of the wall (m)

Calculating Each Parameter

1. **Surface Area (A)**: Each box has a surface area of 6 sides, each side being 1 m². However, since we are interested in the heat transfer through the walls of the middle box, we will consider only the sides exposed to the air. The middle box has 5 sides exposed (top, bottom, and 4 sides). Thus:

A = 5 × (1 m × 1 m) = 5 m²

2. **Temperature Difference (ΔT)**: The temperature difference between the inside of the box (0 °C) and the outside air (30 °C) is:

ΔT = 30 °C - 0 °C = 30 K

3. **Wall Thickness (d)**: The thickness of the wall is given as 0.3 cm, which we convert to meters:

d = 0.3 cm = 0.003 m

Substituting Values into Fourier's Law

Now we can substitute these values into Fourier's law to find the thermal conductivity (k):

99900 J/s = k × 5 m² × (30 K) / 0.003 m

Rearranging the equation to solve for k:

k = (99900 J/s × 0.003 m) / (5 m² × 30 K)

Calculating this gives:

k = (299.7) / (150) = 1.998 W/m·K

Final Result

The thermal conductivity of the material of the box is approximately 2.0 W/m·K. This value indicates how well the material conducts heat, which is crucial for understanding how quickly the ice melts in the middle box due to the heat from the surrounding air.