Answer:
In kJ/kg.K - 1.005 kJ/kg degrees Kalvin.
In J/g.°C - 1.005 J/g °C
In kcal/ kg °C 0.240 kcal/kg °C
In Btu/lbm-°F 0.240 Btu/lbm degree F
Step-by-step explanation:
given data:
specific heat of air = 1.005 kJ/kg °C
In kJ/kg.K
1.005 kJ./kg °C = 1.005 kJ/kg degrees Kalvin.
In J/g.°C
[tex]1.005 kJ/kg C \times (1000 J/1 kJ) \times (1kg / 1000 g) = 1.005 J/g °C[/tex]
In kcal/ kg °C
[tex]1.005 kJ/kg C \times (\frac{1 kcal}{4.190 kJ}) = 0.240 kcal/kg C[/tex]
For kJ/kg. °C to Btu/lbm-°F
Need to convert by taking following conversion ,From kJ to Btu, from kg to lbm and from degrees C to F.
[tex]1.005 kJ/kg C \frac{1 Btu}{1.055 kJ} \times \frac{0.453 kg}{1 lbm} \times \frac{(5/9)\ degree C}{ 1\ degree F} = 0.240 Btu/lbm degree F[/tex]
1.005 kJ/kg C = 0.240 Btu/lbm degree F
