Converting specific heat of aluminum

Mar 2009
3
0
The specific of aluminum is .22 cal/g*C
Converting to kcal/kg*C = .22 kcal/kg*C

Converting to BTU/lbs*F = .012 BTU/lbs*F

.22 kcal/ kg*C = 1 BTU/.252 kcal = 1 kg/2.2 lbs = 1 C/((9/5)*1+32) = .22 BTU/ 18.73872 lbs*F and then I divided that out to get .012 BTU/lbs*F

Did I do this right?