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.
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]
6 answers
i did what u wrote above. but did not have the answer right. The answer from the book is 13.9 kJ
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.
kJ
whereas the answer for q2 will be in
J
Please get back if you still get the wrong answer.
why are there two q1 and q2?
wait nvm,
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
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
2 answers