Question:

200 g water is heated from \(40^\circ\mathrm{C}\) to \(60^\circ\mathrm{C}\). Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water \(= 4184\ \mathrm{J/kg/K}\)) :

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With negligible expansion, \(W = P\Delta V \approx 0\), so \(\Delta U = Q = ms\Delta T\).
Updated On: Nov 11, 2026
  • 8.4 kJ
  • 4.2 kJ
  • 16.7 kJ
  • 167.4 kJ
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The Correct Option is C

Solution and Explanation

First law: \(\Delta U = Q - W\). Expansion is ignored, so \(W\approx 0\) and \(\Delta U = Q\).
\(Q = ms\Delta T = 0.2\ \mathrm{kg}\times 4184\ \mathrm{J/kg/K}\times 20\ \mathrm{K} = 16736\ \mathrm{J}\approx 16.7\ \mathrm{kJ}\).
Answer: (3) 16.7 kJ
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