Yahoo Clever wird am 4. Mai 2021 (Eastern Time, Zeitzone US-Ostküste) eingestellt. Ab dem 20. April 2021 (Eastern Time) ist die Website von Yahoo Clever nur noch im reinen Lesemodus verfügbar. Andere Yahoo Produkte oder Dienste oder Ihr Yahoo Account sind von diesen Änderungen nicht betroffen. Auf dieser Hilfeseite finden Sie weitere Informationen zur Einstellung von Yahoo Clever und dazu, wie Sie Ihre Daten herunterladen.
Transfer of heat problem?
Steam at 100°C is bubbled into 0.200 kg of water at 15°C in a calorimeter cup. How much steam will have been added when the water in the cup reaches 62°C? (Ignore the effect of the cup.)
The formula I have says to calculate the mass of the steam by:
mass of the water x heat constant of water x change in water temp / - heat constant of steam x change in steam temp.
I get .52 but that's not right. What am I missing? Please help. Thanks!
Sam, I tried the way you said and I got .15 kg but that's still not right. Could you plug the numbers into the formula you gave so I can make sure I set it up right? Thanks!
2 Antworten
- ?Lv 7vor 9 JahrenBeste Antwort
As steam cools, it first gives up the latent heat of vaporization (or condensation) as the gas turns to liquid water.
this value is about 2260 kJ/kg
Then the liquid will give up its specific heat as it cools from 100C to 62C
the specific heat of water is about 4.186 kJ/kg
If m is the mass of steam required
your equation for the solution will be
0.200(4.186)(62 - 15) = m(2260 + (4.186(100 - 62))
m = 0.016 kg
- vor 9 Jahren
you are missing the heat that will be release when water at 100°C (converted from steam) reaches 62°C.
So your final equation should be,
heat of fusion*mass of steam + heat release for water @ 100 to reach 62 = heat absorbed by water at 15 to reach 62.. Hope u got it.. Do ask me for any further if needed !!
