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Finding the equilibrium temperature of a mixture substances

  • #1

Homework Statement



160g of steam at 100°C is added to 1000 g of ice at 0°C. What is the equilibrium temperature of the mixture? Assume the container has a negligible heat capacity.


Homework Equations



Qloss+Qgain=0;[m*c(delta T)] + [m*c(delta T)] = 0

The Attempt at a Solution



i have the final answer which is 19.3 degree Celsius but i dunno how to get it., i used the formula of Qloss+Qgain=0;[m*c(delta T)] + [m*c(delta T)] = 0. but my final answer is not 19.3.., can you please help me? what formula will i need to use? do i need to consider the latent heat of vaporization and fusion?if so, what should be the formula?


this is my solution;

Qloss(steam) + Qgain (ice) = 0

m*C (ΔT) + m*C (ΔT) = 0
160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0
76.8 (t-100) + 500 (t-0) = 0
76.8 t - 7680 + 500 t = 0
576.8 t / 576.8 = 7680/576.8
t = 13.3 °C
 

Answers and Replies

  • #2
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do i need to consider the latent heat of vaporization and fusion?
What will happen to ice exactly at freezing if you add heat to it? What will happen to steam exactly at boiling if you remove heat from it?
 
  • #3
the ice will melt and the steam will evaporate?
 
  • #4
Curious3141
Homework Helper
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What are the distinct processes going on here? Think steam condensation, ice melting and heating of water.

Then take it in stages. First figure out how much heat is generated if all the steam condenses to water at 100 deg C. Then figure out how much heat is absorbed if all the ice melts to water at 0 deg C. There will be an "excess" of heat here, which you should compute and use later.

Now you have some water (how much?) at 100 deg C and some water (how much?) at 0 deg C. Figure out the equilibrium temperature from simply mixing these, which can be done by simple proportion.

Finally, use the heat "excess" you got in the first part to work out how many kelvin (or deg C) the water will go up by, add that to the previous equilibrium temperature you derived after the "mixing" step, and that's your final equilibrium temperature.
 
  • #5
NascentOxygen
Staff Emeritus
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this is my solution;

Qloss(steam) + Qgain (ice) = 0

m*C (ΔT) + m*C (ΔT) = 0
160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0
76.8 (t-100) + 500 (t-0) = 0
76.8 t - 7680 + 500 t = 0
576.8 t / 576.8 = 7680/576.8
t = 13.3 °C
You won't reach a solution until you involve all 3 specific heats: that of steam, of ice, and of water. You have involved only two. The steps Curious3141 detailed should be followed.
 
  • #6
"how much heat is generated"

you mean the specific heat of a substance?
 
  • #7
NascentOxygen
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"how much heat is generated"

you mean the specific heat of a substance?
A better word is "liberated". How much excess heat is liberated when that steam undergoes a phase change?
 
  • #8
where will i place the specific heat of water? since no masses are given for water.

what should be the equation? :(((
 
  • #9
NascentOxygen
Staff Emeritus
Science Advisor
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You're not told how much water there is, no. And it's clear there is none to start with. But if you think about the problem, you'll see where the water is going to come from...

NO! The steam does not "evaporate". The two substances are mixed together in a sealed flask. No mass and no heat is lost.
 
  • #10
still i didn't get it., :((( it's really hard to understand.., i really dunno what formula will i use.., i tried this formula, but still i get the wrong answer., dunno if what's the right value to substitute.. i dunno where to get or which one will i derive to get the right formula.., :'((( *sigh*

Qsteam =m*C (ΔT) + mL + m*C (ΔT) = Qice = -[mL + m*C (ΔT)]
 
  • #11
NascentOxygen
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When equilibrium is reached, you are going to end up with your flask containing a mass M of water at temperature T.

It's time to take a step back, and start over.

I suggest that you set yourself the task of writing a descriptive paragraph explaining in precise detail how the container contents change from what you start with, to what you end up with. In not less than 300 words. By forcing yourself to analyze the situation step by step with sufficient clarity that you are able to make a cogent written description, you will in the process map out the method for solving it mathematically. Forget equations at this stage, if you haven't grasped what's going on, how can you be confident that you'll come up with the right equations?
 
  • #12
i got it! :)))

first, steam to water ; Q=m*Lv : Q= 160(540)

Q=86,400 (dunno the unit) lol. :)))


second, ice to water ; Q=m*Lf : Q=1000(79.6)

Q= 79,600

the excess heat is; (Qsteam) 86,400 - (Qice) 79,600 = 8, 600


then.., the initial temp of both subs.;

Qloss(steam) + Qgain (ice) = 0

m*C (ΔT) + m*C (ΔT) = 0
160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0
76.8 (t-100) + 500 (t-0) = 0
76.8 t - 7680 + 500 t = 0
576.8 t / 576.8 = 7680/576.8
t = 13.3 °C

then, the final temp of both subs. using this formula;

Q = m*c (T.final - T.initial)

where
Q= excess heat
m=mass of steam + mass of ice
c=specific heat of water
T.initial=13.3


Q=1160 (1) (T - 13.3)
6800 =1160T - 15428
6800+15428 = 1160T
22228/1160 = 1160T/1160
19.16 or 19.2 = T


:))) but still not the exact answer.., why? heheh
 
  • #13
6,054
390
then.., the initial temp of both subs.;
What "both subs" do you have here?
 
  • #14
steam and ice?
 
  • #15
Curious3141
Homework Helper
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A few problems.

You seem to be using calorie as the heat unit, which is acceptable, although SI units like J (Joule) are preferred. But this depends on what's covered in your course.

i got it! :)))

first, steam to water ; Q=m*Lv : Q= 160(540)

Q=86,400 (dunno the unit) lol. :)))
Calories!!


second, ice to water ; Q=m*Lf : Q=1000(79.6)

Q= 79,600

the excess heat is; (Qsteam) 86,400 - (Qice) 79,600 = 8, 600
Typo; I guess you meant 6,800 cal (units are important!).

then.., the initial temp of both subs.;

Qloss(steam) + Qgain (ice) = 0
Inaccurate. At this point both the steam and ice have become water, so they are the same substance. You can distinguish them by naming them Q1 and Q2, for instance, but you can't call them steam and ice still.

m*C (ΔT) + m*C (ΔT) = 0
160g*0.48cal/g C° (t-100°C) + 1000g*0.50cal/g C°(t-0) = 0
This shows a misconception. You used different specific heat capacities (those of steam and ice). They're close, so it doesn't affect the answer much, but you should just use one specific heat capacity - that of water, since everything is water at this stage.

76.8 (t-100) + 500 (t-0) = 0
76.8 t - 7680 + 500 t = 0
576.8 t / 576.8 = 7680/576.8
t = 13.3 °C

then, the final temp of both subs. using this formula;

Q = m*c (T.final - T.initial)

where
Q= excess heat
m=mass of steam + mass of ice
c=specific heat of water
T.initial=13.3


Q=1160 (1) (T - 13.3)
6800 =1160T - 15428
6800+15428 = 1160T
22228/1160 = 1160T/1160
19.16 or 19.2 = T


:))) but still not the exact answer.., why? heheh
The rest looks OK. The answer may not match the book's exactly because you may use slightly different values for the specific heats, etc., but you have definitely messed up in the calculations, as I've indicated. Correct that, and see if you have a closer match.
 
  • #16
6,054
390
steam and ice?
How is that possible after you did

first, steam to water ; Q=m*Lv : Q= 160(540)

Q=86,400 (dunno the unit) lol. :)))


second, ice to water ; Q=m*Lf : Q=1000(79.6)

Q= 79,600
?
 
  • #17
ok., thanks a lot! :)))
 

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