One face of the a cube of side 0.2 m is in contact with ice and the opposite face is in contact with steam. If all other sides are well lagged, calculate the mass of ice that melts during one hour. Thermal conductivity of the metal `=40 "Wm"^(-1) "K"^(-1)`. L.H. of fusion of ice `=336 "kJ kg"^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Side of the cube, \( L = 0.2 \, \text{m} \)
- Thermal conductivity of the metal, \( k = 40 \, \text{W m}^{-1} \text{K}^{-1} \)
- Temperature of steam, \( T_1 = 100 \, \text{°C} \)
- Temperature of ice, \( T_2 = 0 \, \text{°C} \)
- Latent heat of fusion of ice, \( L_f = 336 \, \text{kJ/kg} = 336 \times 10^3 \, \text{J/kg} \)
- Time, \( t = 1 \, \text{hour} = 3600 \, \text{s} \)

### Step 2: Calculate the temperature difference
\[ \Delta T = T_1 - T_2 = 100 \, \text{°C} - 0 \, \text{°C} = 100 \, \text{K} \]

### Step 3: Calculate the area of one face of the cube
\[ A = L^2 = (0.2 \, \text{m})^2 = 0.04 \, \text{m}^2 \]

### Step 4: Calculate the rate of heat flow (Q/t)
Using the formula for heat conduction:
\[ \frac{Q}{t} = \frac{k \cdot A \cdot \Delta T}{L} \]
Substituting the values:
\[ \frac{Q}{t} = \frac{40 \, \text{W m}^{-1} \text{K}^{-1} \cdot 0.04 \, \text{m}^2 \cdot 100 \, \text{K}}{0.2 \, \text{m}} \]

### Step 5: Simplify the equation
Calculating the numerator:
\[ = 40 \cdot 0.04 \cdot 100 = 160 \, \text{W} \]
Now divide by the length:
\[ \frac{Q}{t} = \frac{160}{0.2} = 800 \, \text{W} \]

### Step 6: Calculate the total heat transferred in one hour
\[ Q = \frac{Q}{t} \cdot t = 800 \, \text{W} \cdot 3600 \, \text{s} = 2880000 \, \text{J} \]

### Step 7: Relate the heat transferred to the mass of ice melted
Using the latent heat of fusion:
\[ Q = m \cdot L_f \]
Where \( m \) is the mass of ice melted. Rearranging gives:
\[ m = \frac{Q}{L_f} = \frac{2880000 \, \text{J}}{336000 \, \text{J/kg}} = \frac{2880000}{336000} \approx 8.571 \, \text{kg} \]

### Final Answer
The mass of ice that melts during one hour is approximately \( 8.571 \, \text{kg} \).

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