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
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 \;kgV1=1m3p1=344.8kpaT1=273KT2=600Kp1V1=mRT1R=0.287kJ/kgK344.8×1=m×0.287×273m=4.4007kg
The specific heat at constant volume:
cv=0.718 kJ/kg Kc_v = 0.718 \;kJ/kg \;Kcv=0.718kJ/kgK
Internal energy \:
ΔU=mcvΔT=4.4007×0.718×(T2−T1)=3.1597×(600−273)=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 \;kJΔU=mcvΔT=4.4007×0.718×(T2−T1)=3.1597×(600−273)=3.1597×327=1033.2kJ
Answer: 1033 kJ
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