Question

A coil of wire 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 s is :

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

Hint:

Always convert kg to g carefully: \(1\,kg = 1000\,g\).

Answer 1

The Correct Option is B

Concept: \[ P = \frac{V^2}{R}, \quad Q = Pt, \quad Q = mL \]

Step 1: Power.
\[ P = \frac{210^2}{50} = \frac{44100}{50} = 882 \, W \]

Step 2: Heat in 1 s.
\[ Q = 882 \, J \]

Step 3: Mass melted.
\[ L = 3.36 \times 10^5 \, J/kg \] \[ m = \frac{882}{3.36 \times 10^5} = 2.625 \times 10^{-3} \, kg \] \[ = 2.625 \, g \approx 2.62 \, g \]