A cubical ice box of side 50 cm has a thickness of 5.0 cm. if 5 kg of ice is put in the box, estimate the amount of ice remaining after 4 hours. The outside temperature is `40^(@)C` and coefficient of thermal conductivity of the material of the box = `0.01 Js^(-1) m^(-1) .^(@)C^(-1)`. Heat of fusion of ice` = 335 Jg^(-1)`.

Options

A. `3.7 kg`
B. `3.9 kg`
C. `4.7 kg`
D. `4.9 kg`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the dimensions and constants to SI units
- The side length of the ice box \( L = 50 \, \text{cm} = 0.5 \, \text{m} \)
- The thickness of the box \( \Delta x = 5 \, \text{cm} = 0.05 \, \text{m} \)
- The outside temperature \( T_1 = 40 \, \text{°C} \)
- The inside temperature \( T_2 = 0 \, \text{°C} \) (temperature of ice)
- The coefficient of thermal conductivity \( K = 0.01 \, \text{Js}^{-1}\text{m}^{-1}\text{°C}^{-1} \)
- The heat of fusion of ice \( L_f = 335 \, \text{J/g} = 335000 \, \text{J/kg} \)
- The mass of ice \( m_{\text{initial}} = 5 \, \text{kg} \)
- The time \( t = 4 \, \text{hours} = 4 \times 3600 \, \text{s} = 14400 \, \text{s} \)

### Step 2: Calculate the surface area of the ice box
The ice box is cubical, and the surface area \( A \) of the six faces is given by:
\[ A = 6L^2 = 6 \times (0.5 \, \text{m})^2 = 6 \times 0.25 \, \text{m}^2 = 1.5 \, \text{m}^2 \]

### Step 3: Calculate the temperature difference
The temperature difference \( \Delta T \) between the outside and inside of the box is:
\[ \Delta T = T_1 - T_2 = 40 \, \text{°C} - 0 \, \text{°C} = 40 \, \text{°C} \]

### Step 4: Calculate the amount of heat transferred to the ice box
Using the formula for heat transfer:
\[ Q = \frac{K \cdot A \cdot \Delta T \cdot t}{\Delta x} \]
Substituting the values:
\[ Q = \frac{0.01 \, \text{Js}^{-1}\text{m}^{-1}\text{°C}^{-1} \cdot 1.5 \, \text{m}^2 \cdot 40 \, \text{°C} \cdot 14400 \, \text{s}}{0.05 \, \text{m}} \]
Calculating \( Q \):
\[ Q = \frac{0.01 \cdot 1.5 \cdot 40 \cdot 14400}{0.05} = \frac{8640}{0.05} = 172800 \, \text{J} \]

### Step 5: Calculate the mass of ice melted
Using the heat of fusion:
\[ Q = m \cdot L_f \]
Rearranging gives:
\[ m = \frac{Q}{L_f} = \frac{172800 \, \text{J}}{335000 \, \text{J/kg}} \approx 0.516 \, \text{kg} \]

### Step 6: Calculate the remaining mass of ice
The remaining mass of ice after 4 hours is:
\[ m_{\text{remaining}} = m_{\text{initial}} - m = 5 \, \text{kg} - 0.516 \, \text{kg} \approx 4.484 \, \text{kg} \]

### Final Answer
The amount of ice remaining after 4 hours is approximately \( 4.48 \, \text{kg} \).

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