Due to cold weather a 1 m water pipe of cross-sectional area `1 cm^2` is filled with ice at ` - 10^@C`. Resistive heating is used to melt the ice. Current of 0.5 A is passed through `4 kOmega` resistance. Assuming that all the heat produced is used for melting, what is the minimum time required ?
(Given latent heat of fusion for water/ice ` = 3.33 xx 10^5 J kg^(-1)`,
specific heat of ice ` = 2 xx 10^3 J kg^(-1)` and
density of ice `=10^3kg//m^3`)

Options

A. 0..353s
B. 3.53s
C. 35.3s
D. 70.6s

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the volume of ice in the pipe
The volume \( V \) of ice can be calculated using the formula:
\[ V = \text{Length} \times \text{Cross-sectional Area} \]
Given:
- Length of the pipe \( L = 1 \, \text{m} \)
- Cross-sectional area \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)

Calculating the volume:
\[ V = 1 \, \text{m} \times 1 \times 10^{-4} \, \text{m}^2 = 1 \times 10^{-4} \, \text{m}^3 \]

### Step 2: Calculate the mass of ice
The mass \( m \) of ice can be calculated using the formula:
\[ m = \text{Density} \times \text{Volume} \]
Given:
- Density of ice \( \rho = 1000 \, \text{kg/m}^3 \)

Calculating the mass:
\[ m = 1000 \, \text{kg/m}^3 \times 1 \times 10^{-4} \, \text{m}^3 = 0.1 \, \text{kg} \]

### Step 3: Calculate heat required to raise the temperature of ice from -10°C to 0°C
The heat required \( Q_1 \) to raise the temperature can be calculated using:
\[ Q_1 = m \cdot c \cdot \Delta T \]
Where:
- \( c = 2000 \, \text{J/kg°C} \) (specific heat of ice)
- \( \Delta T = 10 \, \text{°C} \)

Calculating \( Q_1 \):
\[ Q_1 = 0.1 \, \text{kg} \times 2000 \, \text{J/kg°C} \times 10 \, \text{°C} = 2000 \, \text{J} \]

### Step 4: Calculate heat required to melt the ice at 0°C
The heat required \( Q_2 \) to melt the ice can be calculated using:
\[ Q_2 = m \cdot L_f \]
Where:
- \( L_f = 3.33 \times 10^5 \, \text{J/kg} \) (latent heat of fusion)

Calculating \( Q_2 \):
\[ Q_2 = 0.1 \, \text{kg} \times 3.33 \times 10^5 \, \text{J/kg} = 33300 \, \text{J} \]

### Step 5: Total heat required
The total heat \( Q \) required to melt the ice is:
\[ Q = Q_1 + Q_2 = 2000 \, \text{J} + 33300 \, \text{J} = 35300 \, \text{J} \]

### Step 6: Calculate the heat produced by the resistive heating
The heat produced \( H \) can be calculated using:
\[ H = I^2 R t \]
Where:
- \( I = 0.5 \, \text{A} \)
- \( R = 4 \, \text{k}\Omega = 4000 \, \Omega \)

Setting \( H = Q \):
\[ 0.5^2 \times 4000 \times t = 35300 \]
\[ 0.25 \times 4000 \times t = 35300 \]
\[ 1000t = 35300 \]
\[ t = \frac{35300}{1000} = 35.3 \, \text{s} \]

### Final Answer
The minimum time required to melt the ice is:
\[ \boxed{35.3 \, \text{s}} \]