0.08 kg$0.08\text{ kg}$ of air is heated at constant volume through 5°C$5^{\circ}\text{C}$. The specific heat of air at constant volume is 0.17 kcal/kg°C$0.17\text{ kcal/kg}^{\circ}\text{C}$ and J = 4.18 joule/cal$J = 4.18\text{ joule/cal}$. The change in its internal energy is approximately:
A.318 J$318\text{ J}$
B.298 J$298\text{ J}$
C.284 J$284\text{ J}$
D.142 J$142\text{ J}$
Solution & Explanation
Related Formula
Q = Δ U + W$$Q = \Delta U + W$$
Since the volume is constant, work done W = 0$W = 0$, leading to:
Δ U = Q = m cv Δ T$$\Delta U = Q = m c_v \Delta T$$
Core Logic
Given values:
m = 0.08 kg$m = 0.08\text{ kg}$cv = 0.17 kcal/kg°C = 0.17 × 10³ cal/kg°C$c_v = 0.17\text{ kcal/kg}^{\circ}\text{C} = 0.17 \times 10^3\text{ cal/kg}^{\circ}\text{C}$Δ T = 5°C$\Delta T = 5^{\circ}\text{C}$J = 4.18 J/cal$J = 4.18\text{ J/cal}$
Isochoric processes channel total thermal input strictly into core state configurations, converting raw metric calories directly to standard mechanical Joules via mechanical equivalent factors (J$J$).
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
