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$0.08	ext{ kg}$ air is heated at constant volume through $5^\circ	ext{C}$. The specific heat of air at constant volume is $0.17	ext{ kcal/kg}^\circ	ext{C}$ and $J = 4.18	ext{ joule/cal}$. The change in its internal energy is approximately. A). $318	ext{ J}$ B). $298	ext{ J}$ C). $284	ext{ J}$ D). $142	ext{ J}$

Edvaya · Accepted Answer

The correct answer is $284	ext{ J}$. **Solution:** Since the heating occurs at a constant volume, the work done is zero ($W = 0$). From the first law of thermodynamics, the change in internal energy equals the heat supplied: $\Delta U = Q$. $Q = m c_v \Delta T$ Given $m = 0.08	ext{ kg}$, $c_v = 0.17	ext{ kcal/kg}^\circ	ext{C} = 0.17 	imes 10^3	ext{ cal/kg}^\circ	ext{C}$, and $\Delta T = 5^\circ	ext{C}$. $Q = 0.08 	imes (0.17 	imes 10^3) 	imes 5 = 0.4 	imes 170 = 68	ext{ cal}$. Converting this to Joules: $\Delta U = 68 	imes 4.18	ext{ J/cal} = 284.24	ext{ J} \approx 284	ext{ J}$.

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