NEET UG PHYSICS MCQ: THERMODYNAMICS

1 Easy

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Application +4 | -1

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 \text{ joule}/ \mathrm{cal}.

The change in its internal energy is approximately

View answer and explanation

Correct option: A

284 J

The change in internal energy is calculated using the formula

\Delta U = mc_v\Delta T

. Substituting the given values, we get

\Delta U = 0.08 \times 0.17 \times 5 \times 4186.8 \approx 284 J

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