A `5 cm` thick ice block is there on the surface of water in a lake. The tmeperature of air `- 10^(@)C`, how muct time it will take to double the thickness of the block?
`(L = 80 cal//g, K_(ice) = 0.004 cal//s-K, d_(ice) = 0.92 g cm^(-3))`

Options

A. `1 hour`
B. `191 hour`
C. `19.1 hour`
D. `1.91 hour`

Explanation

To solve the problem of how long it will take to double the thickness of a 5 cm ice block at -10°C, we will use the formula derived from the principles of heat transfer. The steps are as follows:

### Step 1: Understand the Problem
We need to find the time required to increase the thickness of an ice block from 5 cm to 10 cm, given the temperature of the surrounding air is -10°C.

### Step 2: Identify the Given Data
- Initial thickness of ice, \( d_1 = 5 \, \text{cm} \)
- Final thickness of ice, \( d_2 = 10 \, \text{cm} \) (which means we need to increase the thickness by \( 5 \, \text{cm} \))
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)
- Thermal conductivity of ice, \( K = 0.004 \, \text{cal/s-K} \)
- Density of ice, \( \rho = 0.92 \, \text{g/cm}^3 \)
- Temperature of air, \( T_{\text{air}} = -10^\circ C \)
- Temperature at the bottom of the ice (where it meets water), \( T_{\text{water}} = 0^\circ C \)

### Step 3: Calculate the Temperature Difference
The temperature difference \( \Delta T \) is:
\[ \Delta T = T_{\text{water}} - T_{\text{air}} = 0 - (-10) = 10 \, \text{K} \]

### Step 4: Use the Formula for Time
The formula for the time \( t \) required to increase the thickness of the ice is given by:
\[ t = \frac{\rho \cdot L \cdot \Delta d^2}{2 \cdot K \cdot \Delta T} \]
where:
- \( \Delta d = d_2 - d_1 = 10 \, \text{cm} - 5 \, \text{cm} = 5 \, \text{cm} \)

### Step 5: Substitute the Values
Substituting the known values into the formula:
\[ t = \frac{0.92 \, \text{g/cm}^3 \cdot 80 \, \text{cal/g} \cdot (5 \, \text{cm})^2}{2 \cdot 0.004 \, \text{cal/s-K} \cdot 10 \, \text{K}} \]

Calculating \( (5 \, \text{cm})^2 \):
\[ (5 \, \text{cm})^2 = 25 \, \text{cm}^2 \]

Now substituting this back into the equation:
\[ t = \frac{0.92 \cdot 80 \cdot 25}{2 \cdot 0.004 \cdot 10} \]

### Step 6: Calculate the Numerator and Denominator
Calculating the numerator:
\[ 0.92 \cdot 80 \cdot 25 = 1840 \, \text{cal} \]

Calculating the denominator:
\[ 2 \cdot 0.004 \cdot 10 = 0.08 \, \text{cal/s} \]

### Step 7: Final Calculation
Now substituting the values:
\[ t = \frac{1840}{0.08} = 23000 \, \text{s} \]

### Step 8: Convert Seconds to Hours
To convert seconds to hours:
\[ t_{\text{hours}} = \frac{23000 \, \text{s}}{3600 \, \text{s/h}} \approx 6.39 \, \text{hours} \]

### Final Answer
The time it will take to double the thickness of the ice block is approximately **6.39 hours**.

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