A 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, we need to calculate the quantity of ice that will melt inside the cold box due to heat transfer through the thermocol material. Here’s how we can approach the problem step by step:

### Step 1: Identify the Given Data
- Edge of the cube (L) = 50 cm = 0.50 m
- Thickness of the thermocol (d) = 4 cm = 0.04 m
- Outside temperature (T_out) = 30°C
- Inside temperature (T_in) = 0°C (assuming the inside is kept at freezing point)
- Thermal conductivity of thermocol (K) = 0.050 W/m·K
- Latent heat of fusion of ice (L_f) = 80 cal/g = 80 * 4.2 J/g = 336 J/g = 336,000 J/kg

### Step 2: Calculate the Surface Area of the Cube
The surface area (A) of a cube is given by the formula:
\[ A = 6L^2 \]
Substituting the value of L:
\[ A = 6 \times (0.50)^2 = 6 \times 0.25 = 1.5 \, \text{m}^2 \]

### Step 3: Calculate the Temperature Difference
The temperature difference (ΔT) between the inside and outside is:
\[ \Delta T = T_{out} - T_{in} = 30°C - 0°C = 30°C \]

### Step 4: Use the Heat Transfer Formula
The heat transfer (Q) through the thermocol can be calculated using Fourier’s law of heat conduction:
\[ Q = \frac{K \cdot A \cdot \Delta T}{d} \cdot t \]
Where:
- \( t \) = time in seconds (1 hour = 3600 seconds)

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

### Step 5: Calculate Q
Calculating the above expression step by step:
1. Calculate the numerator:
\[ 0.050 \cdot 1.5 \cdot 30 = 2.25 \, \text{W} \]
2. Divide by the thickness:
\[ \frac{2.25}{0.04} = 56.25 \, \text{W} \]
3. Multiply by time (in seconds):
\[ Q = 56.25 \cdot 3600 = 202500 \, \text{J} \]

### Step 6: Calculate the Mass of Ice that Melts
Using the latent heat of fusion:
\[ Q = m \cdot L_f \]
Where \( m \) is the mass of ice melted. Rearranging gives:
\[ m = \frac{Q}{L_f} \]
Substituting the values:
\[ m = \frac{202500}{336000} \approx 0.60268 \, \text{kg} \]

### Final Answer
The quantity of ice that will melt each hour inside the cold box is approximately:
\[ \text{Mass of ice melted} \approx 0.603 \, \text{kg} \]