JEE Mains · Physics · STD 11 - 11. thermodynamics
\(0.08 \mathrm{~kg}\) air is heated at constant volume through \(5^{\circ} \mathrm{C}\). The specific heat of air at constant volume is \(0.17 \mathrm{kcal} / \mathrm{kg}^{\circ} \mathrm{C}\) and \(\mathrm{J}=4.18\) joule \(/ \mathrm{cal}\). The change in its internal energy is approximately.
- A \(318 \mathrm{~J}\)
- B \(298 \mathrm{~J}\)
- C \(284 \mathrm{~J}\)
- D \(142 \mathrm{~J}\)
Answer & Solution
Correct Answer
(C) \(284 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{Q}=\Delta \mathrm{U} \text { as work done is zero [constant volume] } \) \( \Delta \mathrm{U}=\mathrm{ms} \Delta \mathrm{T} \) \( =0.08 \times(170 \times 4.18) \times 5 \) \( \simeq 284 \mathrm{~J}\)
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