How to find cp and cv from example problem 11.1

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williamcarter
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Hello, I tried attempting the following example problem 11.1(please see below), however I don't quite get from where they got cp=1.00 and cv=0.717?
I know that Q=m*cp*delta T=> cp=Q/(m*delta T)
Q=n*cv*delta T=>cv=Q/(n*deltaT)

Problem statement
Textbook page showing Example Problem 11.1: air flows steadily through a constant-area duct cooled by liquid nitrogen, with inlet T1 = 440 K, p1 = 188 kPa (abs), V1 = 210 m/s and outlet T2 = 351 K, p2 = 213 kPa (abs), and a schematic of the duct with heat transfer Q-dot < 0.
Their solution:
Worked solution for finding duct area and changes in enthalpy and internal energy of an ideal gas, using continuity and cp = 1.00 kJ/kg·K, cv = 0.717 kJ/kg·K with temperatures 351 K and 440 K.


How did they manage to get for cp=1.00 and cv=0.717?

Thank you
 

Discussion

Chestermiller said:
(3.5)(8.314)/29=1.00
Thank you , but how did you get these numbers from? They are not given in the statement of the problem.
What about cv?
 
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