Asked by Janus
5 pounds of an ideal gas with R=38.7 and K=1.668 have 300Btu of heat added during a reversible constant pressure change of state, initial temperature=80°F.
Determine:
a.)the final temperature
b.)change of internal energy
change of enthalpy
change of entropy
Determine:
a.)the final temperature
b.)change of internal energy
change of enthalpy
change of entropy
Answers
Answered by
drwls
a) Since it is a constant pressure heating, dQ = M Cp dT = [K*M*R/((K-1)]*dT
Use that to determine the temperature changle dT, and add that to 80 F.
You need to specify the units of R. They should be Btu/lbm*degF
b) dU = M Cv dT = [M/(K-1)]dT
c) dH = Cp*dT = dQ
d) dS = Integral of dQ/T
= Integral of M Cp dT/T
= M Cp ln(T2/T1)
The T's must be in degrees Rankine
Use that to determine the temperature changle dT, and add that to 80 F.
You need to specify the units of R. They should be Btu/lbm*degF
b) dU = M Cv dT = [M/(K-1)]dT
c) dH = Cp*dT = dQ
d) dS = Integral of dQ/T
= Integral of M Cp dT/T
= M Cp ln(T2/T1)
The T's must be in degrees Rankine
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