Question

The mass defect in a particular reaction is \( 0.4 \, \text{g} \). The amount of energy liberated is \( n \times 10^7 \, \text{kWh} \), where \( n = \) _____. (speed of light \( = 3 \times 10^8 \, \text{m/s} \))

Answer 1

Correct Answer: 1

Step 1: Energy Calculation via \( E = \Delta mc^2 \)

Energy \( E \) is determined using Einstein's equation \( E = \Delta mc^2 \), with the values:

\[ E = 0.4 \times 10^{-3} \times (3 \times 10^8)^2. \]

The simplified result is:

\[ E = 3600 \times 10^7 \, \text{kWs}. \]

Step 2: Conversion to kWh

To convert kilowatt-seconds (kWs) to kilowatt-hours (kWh), divide by 3600 (seconds per hour):

\[ \frac{3600 \times 10^7}{3600} \, \text{kWh} = 1 \times 10^7 \, \text{kWh}. \]