Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 4 (Q.No. 5)
5.
1 Kcal/kg. °C is equivalent to __________ BTU/lb. °F.
Discussion:
1 comments Page 1 of 1.
SURADA YERRAYYA said:
2 years ago
1 BTU = 252 cal ==>1 BTU = 0.252 kcal ==> 1 kcal = 1/0.252 BTU,
1 lb = 0.4535 kg ==> 1 kg =1/0.4535 lb.
1 deg C = 9/5 deg F.
==> (1/0.252)/ ((1/0.4535 ) *(9/5 )).
==> (0.4535*5)/(0252*9) =1.
The answer = A.
1 lb = 0.4535 kg ==> 1 kg =1/0.4535 lb.
1 deg C = 9/5 deg F.
==> (1/0.252)/ ((1/0.4535 ) *(9/5 )).
==> (0.4535*5)/(0252*9) =1.
The answer = A.
(1)
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