Question

A body of specific heat \(0.2\ \text{kcal/kg}^\circ\text{C}\) is heated through \(100^\circ\text{C}\). The percentage increase in its mass is

• A. \(9%\)
• B. \(9.3 \times 10^{-11}%\)
• C. \(10%\)
• D. None of these

Hint:

Mass increase is extremely small because \(c^2\) is huge.

Answer 1

The Correct Option is B

To determine the percentage increase in mass when a body is heated, we can use the relativistic mass-energy equivalence principle. According to Einstein's mass-energy equivalence principle, the energy added to a body also increases its mass. This principle can be expressed using the famous equation:

\(E = mc^2\)

Where:

• \(E\) is the energy added to the body.
• \(m\) is the increase in mass.
• \(c\) is the speed of light in vacuum, approximately \(3 \times 10^8 \text{ m/s}\).

First, calculate the energy added to the body when it is heated through \(100^\circ\text{C}\). This can be calculated using the formula:

\(\Delta Q = mc\Delta T\)

Given data:

• Specific heat capacity, \(c = 0.2 \text{ kcal/kg}^\circ\text{C}\)
• Temperature change, \(\Delta T = 100^\circ\text{C}\)

Therefore, the energy added in calories is:

\(\Delta Q = m \times 0.2 \times 100\)

\(= 20m \text{ kcal}\)

Converting this energy to joules (since \(1 \text{ kcal} = 4184 \text{ J}\)):

\(\Delta Q = 20m \times 4184 \text{ J}\)

Now, use \(E = mc^2\) to find the increase in mass:

\(m_{\text{increase}} = \frac{\Delta Q}{c^2}\)

\(= \frac{20m \times 4184}{(3 \times 10^8)^2}\)

Calculate the fraction of mass increase:

\(= \frac{20 \times 4184}{9 \times 10^{16}}\)

Therefore, the correct answer is \(9.3 \times 10^{-11}%\), which matches option:

\(9.3 \times 10^{-11}%\)