Solveeit Logo

Question

Physics Question on Thermodynamics

0.08 kg of air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal/kg°C and J=4.18J = 4.18 joule/cal. The change in its internal energy is approximately:

A

318 J

B

298 J

C

284 J

D

142 J

Answer

284 J

Explanation

Solution

Since the process is at constant volume, the change in internal energy ΔU\Delta U is given by:

ΔU=msΔT\Delta U = ms\Delta T

where m=0.08kgm = 0.08 \, \text{kg}, s=0.17kcal/kgCs = 0.17 \, \text{kcal/kg}^\circ\text{C}, and ΔT=5C\Delta T = 5^\circ \text{C}.

Convert ss from kcal to joules:

s=0.17×1000×4.18J/kgCs = 0.17 \times 1000 \times 4.18 \, \text{J/kg}^\circ\text{C}

Then,

ΔU=0.08×(0.17×1000×4.18)×5284J\Delta U = 0.08 \times (0.17 \times 1000 \times 4.18) \times 5 \approx 284 \, \text{J}