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

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

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

`1.005 kJ/kg C * (1000 J/1 kJ) * (1kg / 1000 g) = 1.005 J/g °C`

In kcal/ kg °C

`1.005 kJ/kg C * ((1 kcal)/(4.190 kJ)) = 0.240 kcal/kg C`

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.

`1.005 kJ/kg C (1 Btu)/(1.055 kJ) * (0.453 kg)/(1 lbm) * ((5/9)\ degree C)/( 1\ degree F)  = 0.240 Btu/lbm degree F`

1.005 kJ/kg C = 0.240 Btu/lbm degree F