Keep the Heat B/Thermodynamics C

[ChrisYim]

Hey guys. I just wanted to clarify my interpretation of the rules.
We attach an insulator directly on top of the beaker as long as there's a 1.5 cm hole in the center, right?

[135scioly]

Hey guys. I just wanted to clarify my interpretation of the rules.
We attach an insulator directly on top of the beaker as long as there's a 1.5 cm hole in the center, right?

Yes, you can have an insulator on top but it shouldn't be too thick because keep in mind that your beaker must be 2.5 cm from the top of your insulator, which some people interpret to be the tallest point of your box.

Also, can somebody explain the boundary work equation for constant pressure and temperature?

[Luo]

the top of your insulator, which some people interpret to be the tallest point of your box.

This has also been made explicit by an FAQ/rules clarification: http://soinc.org/node/1098

[arod129]

One question, is cork allowed or not?

[siciscio]

Cork is allowed, why wouldn't it be? (not like it's commercial insulation or something )

[135scioly]

Does anyone know the answer to question 18 on the Conestoga practice test?

The question is:

How much thermal energy is required to melt 10 g of ice at -20 degrees Celsius to change it all to steam at 120 degrees celsius?

Thanks

[Skink]

Someone else can jump in if I made an error or did it incorrectly.
It's a stepwise heating problem using heats of fusion and vaporization.

10g ice, 10g water

-20deg C-->0deg C
Q=smΔT (note: have the specific heats of the different phases of water in your binder, too…I made an error assuming ice was 4.184 from memory. It isn’t. And find a more accurate value than I did, preferably from a reliable source)
Q=(2.03J/gdegC)*(10g)*(20degC)
Q=406J

Now, we need to look up the energy required to melt the ice (because, recall, the phase change soaks up a certain amount of energy). Have it in your binder.
Heat of Fusion (Water), 334J/g
(334J/g)*(10g) = 3340J

0degC-->100degC
Q=smΔT
Q=(4.184J/gdegC)*(10g)*(100degC)
Q=4184J

Heat of Vaporization (Water), 2257J/g
(2257J/g)*(10g) = 22570J

100deg C-->120degC
Q=smΔT
Q=(2.080J/gdegC)*(10g)*(20degC)
Q=416J

Add ‘em up.
Qtot = 406J + 3340J + 4184J + 22570J + 416J = 30916J = 30.916kJ
sig figs
30.9kJ

[135scioly]

Thank you soo much! That was really helpful!

[sjwon3789]

Can people define "Commercial Insulators?" I'm really confused with what insulators are prohibited or not because isn't denim a commercial? I think I saw a team bring that to invitationals or I would just be misunderstanding.

[Skink]

There's a FAQ for that, actually, here.