A coil of water of resistance `50 Omega` is embedded in a block of ice and a potential difference of 210 V is applied across it. The amount of ice which melts in 1 second is [ latent heat of fusion of ice `= 80 cal g^(-1)`]

Options

A. 0.262g
B. 2.62g
C. 26.2g
D. 0.0262g

Answer

To solve the problem, we need to determine the amount of ice that melts in 1 second when a potential difference is applied across a coil of resistance. Here are the steps to arrive at the solution:

### Step 1: Calculate the Power (Heat Produced) in the Coil
The power \( P \) produced by the coil can be calculated using the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the potential difference and \( R \) is the resistance.

Given:
- \( V = 210 \, \text{V} \)
- \( R = 50 \, \Omega \)

Substituting the values:
\[ P = \frac{(210)^2}{50} \]

### Step 2: Calculate \( P \)
Calculating \( P \):
\[ P = \frac{44100}{50} = 882 \, \text{W} \]

### Step 3: Calculate the Heat Produced in 1 Second
Since power is defined as energy per unit time, the amount of heat \( Q \) produced in 1 second is equal to the power:
\[ Q = P \times t \]
For \( t = 1 \, \text{s} \):
\[ Q = 882 \, \text{J} \]

### Step 4: Convert Heat to Calories
Next, we need to convert the energy from joules to calories. The conversion factor is:
\[ 1 \, \text{cal} = 4.2 \, \text{J} \]
Thus, to convert joules to calories:
\[ Q_{\text{cal}} = \frac{Q}{4.2} \]
Substituting the value of \( Q \):
\[ Q_{\text{cal}} = \frac{882}{4.2} \approx 210 \, \text{cal} \]

### Step 5: Calculate the Mass of Ice Melted
The amount of ice melted can be calculated using the formula:
\[ m = \frac{Q_{\text{cal}}}{L} \]
where \( L \) is the latent heat of fusion of ice. Given:
- \( L = 80 \, \text{cal/g} \)

Substituting the values:
\[ m = \frac{210 \, \text{cal}}{80 \, \text{cal/g}} = 2.625 \, \text{g} \]

### Final Result
The amount of ice that melts in 1 second is approximately:
\[ m \approx 2.625 \, \text{g} \]

### Summary
Thus, the final answer is:
\[ \text{Amount of ice melted in 1 second} \approx 2.62 \, \text{g} \]