The opposite faces of a cubical block of iron of cross-section `4 xx 10^(-4) "m"^(2)` are kept in contact with steam and melting ice. Calculating the amount of ice melted at the end of 5 minutes. Given `K=0.2` cal cm `""^(-1) "s"^(-1)°"C"^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert Given Quantities to CGS Units
- We are given the cross-sectional area \( A = 4 \times 10^{-4} \, \text{m}^2 \).
- Convert to cm²:
\[ A = 4 \times 10^{-4} \, \text{m}^2 \times \left( \frac{10000 \, \text{cm}^2}{1 \, \text{m}^2} \right) = 4 \, \text{cm}^2 \]
- The length of the cube \( L \) can be derived from the volume of the cube. Since it is cubical, we can take \( L = 2 \, \text{cm} \).

### Step 2: Identify the Temperature Difference
- The temperature difference \( \Delta \theta \) between steam (100°C) and melting ice (0°C) is:
\[ \Delta \theta = 100 - 0 = 100 \, \text{°C} \]

### Step 3: Use the Formula for Heat Conduction
- The rate of heat conduction \( \frac{dQ}{dt} \) is given by:
\[ \frac{dQ}{dt} = \frac{k \cdot A \cdot \Delta \theta}{L} \]
- Substitute the values:
- \( k = 0.2 \, \text{cal} \, \text{cm}^{-1} \, \text{s}^{-1} \, \text{°C}^{-1} \)
- \( A = 4 \, \text{cm}^2 \)
- \( \Delta \theta = 100 \, \text{°C} \)
- \( L = 2 \, \text{cm} \)

### Step 4: Calculate the Rate of Heat Transfer
- Plugging in the values:
\[ \frac{dQ}{dt} = \frac{0.2 \cdot 4 \cdot 100}{2} = \frac{80}{2} = 40 \, \text{cal/s} \]

### Step 5: Calculate the Total Heat Transferred in 5 Minutes
- Convert 5 minutes to seconds:
\[ 5 \, \text{minutes} = 5 \times 60 = 300 \, \text{s} \]
- Total heat \( Q \) transferred in 300 seconds:
\[ Q = \frac{dQ}{dt} \times t = 40 \, \text{cal/s} \times 300 \, \text{s} = 12000 \, \text{cal} \]

### Step 6: Relate Heat to Mass of Ice Melted
- The latent heat of fusion of ice \( L_f \) is given as \( 80 \, \text{cal/g} \).
- The mass \( m \) of ice melted can be calculated using:
\[ Q = L_f \cdot m \implies m = \frac{Q}{L_f} \]
- Substitute the values:
\[ m = \frac{12000 \, \text{cal}}{80 \, \text{cal/g}} = 150 \, \text{g} \]

### Final Answer
The amount of ice melted at the end of 5 minutes is **150 grams**.

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