• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

Thermodynamics Help

J

jmcoreymv

Diamond Member
Im working on the following problem.

If i put 50 grams of steam at 120 celsius into an isolated container with 100 grams of ice at -10 celsius and the system comes to equilibrium, assume the container has a mass of 50 gm and a heat capacity of .2 cal/gm-deg. Assume no losses to the outside encironment. Assume the specific heat of the capacities of both ice and steam are half that of water. Let c(water) = 1 cal/gm-deg, L(ice)=80 cal/gm, L(steam)=540 cal/gm.

Write the energy balance equation that expresses this system coming to equilibrium and identify each term in the equation. Determine the final temp of the system when it reaches equilibrium. Is the final equilibrium state of the system ice and water, all water, or water and steam? Determine how many grams of water there are when it all comes to equilibrium.

I worked it out but I know I totally screwed up because my final temperature was about 8100 deg C...which is obviously just a tad too high. Heres what I tried:

1. Found Q needed to bring the ice and cup to 0 deg C which I calculated as 600 cal.
2. Found Q needed to turn the ice to water which is 8000 cal.
3. Found Q available if steam is brought down to 100 deg C which is 500 cal.
4. Found Q available if all steam at 100 deg C is converted to water at 100 deg C which is 27000 cal.

So in order for the ice to turn to water (at 0 deg C) it will need 8600 cal. So it looks like the final mixture will be a combination of water and steam. But Im not quite sure how to go from here. Any help appreciated.

In case anyone forgot the equations are Q=m*c*delta T for temperature change and Q=mL for latent heat.
 
Here is cliffnotes
1. 100 gm of ice at -10 c in 50 gm cup at -10 c
2. Add 50 gm of steam at 120 c and let it goto equilibrium.
Let c(water) = 1 cal/gm-deg, L(ice)=80 cal/gm, L(steam)=540 cal/gm.
c(steam) and c(ice) = .5
Whats the final state of whats inside? What temperature is it at? How much water is there?
 
this is basic phy, you need to find the temp the ice melt down to water and add the ice water to the water started with, then use Q=mcat

edit: Q=MCAT
A as in delta
 
My guess is that the final temp will be 100 deg C, but how do I figure out how much of that is water and how much is steam?
 
well, one way to think of it as, you got some amount of energy stored in the ice and steam. well total it all out average per gm and figure out what the energy stored in water/steam mixture. keep in mind they will be at the same temp ie 100C if you're correct in a steam/water mixture.

the other way is your method, subtract and add. you've converted your ice to water@100C 8600cal.

and the steam to steam @100C is 500 cal.

so you need to still make up 8100 cal, so that's about 1/3 so that's about 15g steam converted to water@100c.

also i'm not sure if you had to calculate the energy needed to heat up that container.
 
heat flow out of the steam: Qs=MsCs(T-Ts)
heat flow into the water...: Qw=MwCw(T-Tw)
apply conservation of E.....: Qw+Qa=0
solve for T
 
assume the container has a mass of 50 gm and a heat capacity of .2 cal/gm-deg. Assume no losses to the outside encironment.
???
 
It's been forever since I did this stuff. That's my disclaimer.


I believe you're right, you're going to have a mix of water and steam, both at 100. That 27000 calories is a whopping big amount of energy compared to what is required to actually heat the ice, cup, and resulting cold water up to 100 C. So all of the steam won't convert to water, just some of it. We therefore need to solve for the mass of steam that will convert to water.

Q + W = delta U

Q steam + Q steam vaporization - Q ice - Q cup - Q ice melting - Q cold water = 0

(using your calculations below where applicable)

500 + m*L - 600 - (50*0.2*100) - 8000 - (100*1*100) = 0

solve for m


Sound reasonable? Hope this is right.
 
I love thermo, and it appears the you're on the right path:

At the point you left off... you have 27500-8600 = 18900 cals.

1) Take the water at 0C to 100C, requires Q1 (leaving you with 18900cal - Q1)
2) Take the answer from above (18900cal - Q1), this energy will change an amount of water to steam (m = Q/L = (18900cal - Q1)/540 )

Crypticburn


 
Man you're lucky. In my AP Physics C class we are doing banked curves. This is the only topic I do not enjoy for some reason.
 
Originally posted by: jmcoreymv
Banked curves of what? I took Phys C, it doesnt cover any thermo.
Click to expand...

Oh no, it has nothing to do with Thermo. I'm just saying that is the topic we are on. We did thermo in Physics B last year.
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top