q=sm(final-initial temperature).
s= 4.18
m=5.40g
How much heat would be relaesed by the condensation of 5.40g of steam at 100 (degrees)C and the subsequent cooling of the water to 25(degrees)C? [(delta)Hvap=40.7 KJ/mol at 100(degrees)C; Cp for H2O(l) is 4.18Jg-1 (degrees)c-1]
A) 12.2 kj
B) 18.3 kj
C) 12.8 kj
D) 13.9 kj
E) 23.7 kj
2 answers
q1 to condense steam at 100 C.
q1 = mass x heat vap
q2 = heat released cooling from 100 C to 25 C.
q2 = mass x specific heat x (Tfinal-Tinitial)
Total q = q1 + q2.
q1 = mass x heat vap
q2 = heat released cooling from 100 C to 25 C.
q2 = mass x specific heat x (Tfinal-Tinitial)
Total q = q1 + q2.
0 answers