If a wire of resistance `20 Omega` is covered with ice and a voltage of `210 V` is applied across the wire , then the rate of melting of ice is

Options

A. `8.85 gs^(-1)`
B. `1.92 gs^(-1)`
C. `6.56 gs^(-1)`
D. none of these

Explanation

To solve the problem of determining the rate of melting of ice when a wire of resistance \(20 \, \Omega\) is covered with ice and a voltage of \(210 \, V\) is applied across the wire, we can follow these steps:

### Step 1: Calculate the Current (I) through the Wire
Using Ohm's Law, we can find the current flowing through the wire:
\[ I = \frac{V}{R} \]
Where:
- \(V = 210 \, V\) (voltage applied)
- \(R = 20 \, \Omega\) (resistance of the wire)

Substituting the values:
\[ I = \frac{210 \, V}{20 \, \Omega} = 10.5 \, A \]

### Step 2: Calculate the Power (P) Dissipated in the Wire
The power dissipated in the wire can be calculated using the formula:
\[ P = I^2 R \]
Substituting the values:
\[ P = (10.5 \, A)^2 \times 20 \, \Omega = 110.25 \times 20 = 2205 \, W \]

### Step 3: Determine the Rate of Heat Generation
The rate of heat generation (which is equal to the power calculated) is:
\[ \text{Rate of heat generation} = P = 2205 \, J/s \]

### Step 4: Calculate the Rate of Melting of Ice
The rate of melting of ice can be calculated using the latent heat of fusion of ice. The latent heat \(L\) of ice is approximately \(336 \, J/g\). The mass of ice melted per second can be calculated using the formula:
\[ \frac{m}{t} = \frac{P}{L} \]
Where:
- \(P = 2205 \, J/s\) (power)
- \(L = 336 \, J/g\) (latent heat of fusion)

Substituting the values:
\[ \frac{m}{t} = \frac{2205 \, J/s}{336 \, J/g} \approx 6.5625 \, g/s \]

### Final Answer
The rate of melting of ice is approximately \(6.56 \, g/s\).

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