‘cold box’ in the shape of a cube of edge 50 cm is made of ‘thermocol’ material 4.0 cm thick. If the outside temperature is `30^(@)C` the quantity of ice that will melt each hour inside the ‘cold box’ is : (the thermal conductivity of the thermocol material is 0.050 `W//mK)`

Options

A. 1.52 kg
B. 0.605 kg
C. 2.520 kg
D. 0.512 kg

Explanation

To solve the problem of how much ice will melt in the cold box made of thermocol, we will follow these steps:

### Step 1: Understand the Problem
We have a cold box in the shape of a cube with:
- Edge length = 50 cm
- Thickness of thermocol = 4 cm
- Outside temperature = 30°C
- Thermal conductivity of thermocol (K) = 0.050 W/mK
- Latent heat of fusion of ice (L) = 80 cal/g = 334,000 J/kg (after conversion)

### Step 2: Convert Units
Convert the dimensions from centimeters to meters:
- Edge length of the cube = 50 cm = 0.5 m
- Thickness of thermocol = 4 cm = 0.04 m

### Step 3: Calculate the Area
The total surface area (A) of the cube can be calculated as:
\[ A = 6 \times (\text{side length})^2 \]
\[ A = 6 \times (0.5 \, \text{m})^2 = 6 \times 0.25 = 1.5 \, \text{m}^2 \]

### Step 4: Determine the Temperature Difference
The temperature difference (ΔT) between the inside and outside of the box is:
\[ \Delta T = T_{\text{outside}} - T_{\text{inside}} = 30°C - 0°C = 30 \, \text{K} \]

### Step 5: Calculate the Heat Transfer
Using Fourier's law of heat conduction, the heat transfer (Q) through the thermocol in one hour can be calculated as:
\[ Q = \frac{K \cdot A \cdot \Delta T}{\Delta x} \cdot t \]
Where:
- \( K = 0.050 \, \text{W/mK} \)
- \( A = 1.5 \, \text{m}^2 \)
- \( \Delta T = 30 \, \text{K} \)
- \( \Delta x = 0.04 \, \text{m} \)
- \( t = 3600 \, \text{s} \) (1 hour)

Substituting the values:
\[ Q = \frac{0.050 \cdot 1.5 \cdot 30}{0.04} \cdot 3600 \]

### Step 6: Calculate Q
Calculating the first part:
\[ Q = \frac{0.050 \cdot 1.5 \cdot 30}{0.04} = \frac{2.25}{0.04} = 56.25 \, \text{W} \]
Now, multiplying by time:
\[ Q = 56.25 \cdot 3600 = 202500 \, \text{J} \]

### Step 7: Relate Heat to Mass of Ice
The heat required to melt mass \( m \) of ice is given by:
\[ Q = m \cdot L \]
Where \( L = 334,000 \, \text{J/kg} \) (latent heat of fusion of ice).

### Step 8: Solve for Mass (m)
Rearranging the equation gives:
\[ m = \frac{Q}{L} = \frac{202500}{334000} \]

### Step 9: Calculate m
Calculating the mass:
\[ m \approx 0.605 \, \text{kg} \]

### Final Answer
The quantity of ice that will melt each hour inside the cold box is approximately **0.605 kg**.