Question

How much energy (in kilojoules) is released when 14.7 g of steam at 118.0°C is condensed to give liquid water at 77.0°C? The heat of vaporization of liquid water is 40.67 kJ/mol, and the molar heat capacity is 75.3 J/(K·mol) for the liquid and 33.6 J/(K·mol) for the vapor.

Answer 1

@TuringTest @Zarkon

Answer 2

Multi-step question...it's do-able though. Give me a few minutes..

Answer 3

Haha looks familiar... I'll give you this one @dtan5457 :P

Answer 4

This one confused me a bit because if you were to use the given info in the question... such as "33.6 J/(K·mol) for the vapor", you would need to turn 14.7 grams of h20 into moles, which isn't how I usually do problems like these...

I like to leave the mass as 14.7 grams...
so...

1. Calculate the amount of heat released by cooling the system back to boiling point (100 degrees c)

14.7 g x 2.01(specific heat of h20 as a gas) x 18k=531 J or 0.53KJ

2. Heat released by the phase change of gas to liquid...

14.7 x 2260J( vaporization energy of water)=33222 joules or 33.22 KJ

3. Heat released from cooling the water back down from 100 to 80 degrees c.

14.7 x 4.184 (specific heat of h20 as a liquid) x 20=1230 joules or 1.23 KJ.

add it all up...

1.23KJ+32.22kJ+0.53KJ=33.98 KJ released...

Answer 5

However, if you do use the info given and do turn everything into the unit of moles...I'm pretty sure you'd get the exact same answer.

Answer 6

For example..

the phase change (heat released by condensation)

If you use the info given..

(40.67) kj for heat of vaporization..

14.7 grams is roughly 0.816 moles.

0.816 x 40.67=33.22 KJ, the same answer I got

Answer 7

Opps, let me fix

"1.23KJ+32.22kJ+0.53KJ=33.98 KJ released..."

It's "1.23KJ+33.22kJ+0.53KJ=34.98 KJ released...

Answer 8

thank you! @dtan5457

Answer 9

yw. it's quite a time-consuming question

Answer 10

i would hope i don't get a question like this on a test

Answer 11

haha i hope i dont get this question on my test next tuesday D:

Answer 12

oh.. Also can you help me with one more question?

Answer 13

sure. close this question and make a new one. I'll take a good look at it.

Answer 14

okay