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 using \( E = \Delta mc^2 \) 

The energy \( E \) is calculated using Einstein's famous equation \( E = \Delta mc^2 \), where:

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

Simplifying, we get:

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

Step 2: Converting to kWh

To convert from kilowatt-seconds (kWs) to kilowatt-hours (kWh), we divide by 3600 (the number of seconds in an hour):

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

Answer 2

The energy liberated is given by:

\[ E = \Delta m c^2 \]

Substituting the values:

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

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

Converting to kWh:

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