Question

How much heat would be released by the condensation of 5.40 g of steam at 100 °C and the subsequent cooling of the water to 25 °C? [ÄHvap = 40.7 kJ/mol at 100 °C; Cp for H2O(l) is
4.18 J g-1 °C-1]

Answer 1

Heat is released by condensation steam at 100 C to water at 100 C.
q1 = mass x Hvap.

q2 = heat is released by cooling water from 100 C to 25 C.
q2 = mass x specific heat x (Tf - Ti)
where Tf is final T and Ti is initial T.
Both q1 and q2 are negative numbers.
Total heat released is q1+q2.

Answer 2

i did what u wrote above. but did not have the answer right. The answer from the book is 13.9 kJ

Answer 3

Check your working. Note that dHvap is per MOLE and the answer q1 will be in

kJ

whereas the answer for q2 will be in

J

Please get back if you still get the wrong answer.

Answer 4

why are there two q1 and q2?

Answer 5

wait nvm,

Answer 6

q1 = (5.40g/18.02g/mol)(40.7kJ/mol) = 12.20kJ
q2 = (5.40g)(4.18Jg-1°C-1)(100°C - 25°C) = 1692.9J = 1.6929kJ
q1+q2
12.20kJ + 1.6929kJ = 13.9 kJ