Question

The constant-pressure specific heat of air at 25°C is 1.005 kJ/kg °C. Express this value in kJ/kg-K. J/g °C, kcal/kg-°C, and Btu/lbm-°F.

Answer 1

Answer:

Cp= 1.005 kJ/kg °C =  1.005 kJ/(kg*K) = 1.005 J/g°C = 4.206 J/g°C = 0.776 BTU/lb°F

Explanation:

for the specific heat

1) Cp= 1.005 kJ/kg °C * (1 °C/ 1 K) (temperature differences) = 1.005 kJ/(kg*K)

2) Cp= 1.005 kJ/kg °C * (1000 J/ kJ)* (1 kg/1000 gr) = 1.005 J/g°C

3) Cp= 1.005 kJ/kg °C * (4.186 kcal/kJ) = 4.206 J/g°C

4) Cp= 1.005 kJ/kg °C * (1 BTU/1.055 kJ)* (0.4535 kg/lb)*(1.8°C/°F)= 0.776 BTU/lb°F