An ice box used for keeping eatables cool has a total wall area of `1m^2` and a wall thichness of 5.0 cm. The thermal cunductivity of the ice box is `K=0.01 J//m^@C`. It is filled with large amount of ice at `0^@C` along with eatables on a dfay when the temperature is `30^@C` The latent heat of fusion of ice is `334xx10^3 J//kg`. The amount of ice melted in one day is ( 1 day`=86,000s`)

Options

A. 776 g
B. 7760 g
C. 11520 g
D. 1552 g

Explanation

To solve the problem, we will use the formula for heat transfer through conduction and the relationship between heat, mass, and latent heat of fusion.

### Step 1: Identify the given values
- Total wall area, \( A = 1 \, m^2 \)
- Wall thickness, \( \Delta x = 5.0 \, cm = 0.05 \, m \)
- Thermal conductivity, \( K = 0.01 \, J/(m \cdot °C) \)
- Temperature difference, \( \Delta T = 30°C - 0°C = 30°C \)
- Time, \( t = 1 \, day = 86,400 \, s \)
- Latent heat of fusion of ice, \( L = 334 \times 10^3 \, J/kg \)

### Step 2: Calculate the heat transferred through the wall
The formula for heat transfer through conduction is given by:
\[ Q = \frac{K \cdot A \cdot \Delta T \cdot t}{\Delta x} \]
Substituting the values into the equation:
\[ Q = \frac{0.01 \, J/(m \cdot °C) \cdot 1 \, m^2 \cdot 30 \, °C \cdot 86400 \, s}{0.05 \, m} \]

### Step 3: Simplify the equation
Calculating the numerator:
\[ Q = \frac{0.01 \cdot 1 \cdot 30 \cdot 86400}{0.05} \]
Calculating the values:
\[ Q = \frac{25920}{0.05} = 518400 \, J \]

### Step 4: Relate heat to the mass of ice melted
The heat absorbed by the ice to melt it is given by:
\[ Q = m \cdot L \]
Where \( m \) is the mass of ice melted and \( L \) is the latent heat of fusion. Rearranging for \( m \):
\[ m = \frac{Q}{L} \]
Substituting the values:
\[ m = \frac{518400 \, J}{334 \times 10^3 \, J/kg} \]

### Step 5: Calculate the mass of ice melted
Calculating \( m \):
\[ m = \frac{518400}{334000} \approx 1.552 \, kg \]

### Final Result
The amount of ice melted in one day is approximately \( 1.552 \, kg \) or \( 1552 \, grams \).

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