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 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
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Mass increase is extremely small because \(c^2\) is huge.
Updated On: Apr 23, 2026
- \(9%\)
- \(9.3 \times 10^{-11}%\)
- \(10%\)
- None of these
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Energy given to body increases its mass according to \(E = \Delta m c^2\).
Step 2: Detailed Explanation:
Heat energy supplied = \(mC\Delta\theta = m \times 0.2 \times 100 \times 4.2 \times 10^3 = 8.4 \times 10^4 m\ \text{J}\).
From \(E = \Delta m c^2\): \(\Delta m = \frac{E}{c^2} = \frac{8.4 \times 10^4 m}{(3 \times 10^8)^2} = \frac{8.4 \times 10^4 m}{9 \times 10^{16}} = 9.33 \times 10^{-13} m\).
Percentage increase = \(\frac{\Delta m}{m} \times 100 = 9.33 \times 10^{-11}%\).
Step 3: Final Answer:
Thus, percentage increase = \(9.3 \times 10^{-11}%\).
Energy given to body increases its mass according to \(E = \Delta m c^2\).
Step 2: Detailed Explanation:
Heat energy supplied = \(mC\Delta\theta = m \times 0.2 \times 100 \times 4.2 \times 10^3 = 8.4 \times 10^4 m\ \text{J}\).
From \(E = \Delta m c^2\): \(\Delta m = \frac{E}{c^2} = \frac{8.4 \times 10^4 m}{(3 \times 10^8)^2} = \frac{8.4 \times 10^4 m}{9 \times 10^{16}} = 9.33 \times 10^{-13} m\).
Percentage increase = \(\frac{\Delta m}{m} \times 100 = 9.33 \times 10^{-11}%\).
Step 3: Final Answer:
Thus, percentage increase = \(9.3 \times 10^{-11}%\).
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