Question #246999

A closed rigid container has a volume of 1 m3 and holds air at 344.8 kPa and 273 K. Heat is added until the temperature is 600 K. Determine the change of internal energy


1
Expert's answer
2021-10-05T15:43:54-0400

V1=1  m3p1=344.8  kpaT1=273  KT2=600  Kp1V1=mRT1R=0.287  kJ/kg  K344.8×1=m×0.287×273m=4.4007  kgV_1= 1 \;m^3 \\ p_1 = 344.8 \;kpa \\ T_1 = 273 \;K \\ T_2 = 600 \;K \\ p_1V_1 =mRT_1 \\ R= 0.287 \;kJ/kg\;K \\ 344.8 \times 1 = m \times 0.287 \times 273 \\ m = 4.4007 \;kg

The specific heat at constant volume:

cv=0.718  kJ/kg  Kc_v = 0.718 \;kJ/kg \;K

Internal energy \:

ΔU=mcvΔT=4.4007×0.718×(T2T1)=3.1597×(600273)=3.1597×327=1033.2  kJΔU = mc_vΔT \\ = 4.4007 \times 0.718 \times (T_2-T_1) \\ = 3.1597 \times (600-273) \\ = 3.1597 \times 327 \\ = 1033.2 \;kJ

Answer: 1033 kJ


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS