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:

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 Kelvin.

In  J/g.°C

$1.005 kJ./kg °C \times 1000/1kJ \times (1kg/1000 g) = 1.005J/g °C$

In kcal/ kg °C

$1.005 kJ./kg °C \times \frac{1 kcal}{4.190 kJ} = 0.240 kcal/kg °C$

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.

$1.005 kJ./kg °C \times \frac{1 Btu}{1.055 kJ} \times \frac{0.453 kg}{1 lbm} \times \frac{5/9 degree C}{1 degree F} = 0.240 Btu/ lbm / degree F$

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