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Basic Thermodynamics (calorimeter)

  • Thread starter irishbob
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  • #1
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Homework Statement


A 50.0g copper calorimeter contains 250 g of water at 20 C. how much steam must be condensed into the water if the final temperature of the system is to reach 50 C?


Homework Equations


Q=-mLv where Q is heat energy, m is mass, Lv is latent vaporization constant of water (3.33x10^5 J/kg).
and
Q=mc(Tf-Ti) where Q is heat energy, m is mass, c is specific heat (steam is 0.48 cal/(gC) or 2010 J/(kgC)), and Tf-Ti is delta T.


The Attempt at a Solution


100 C steam -> 100 C water -> 50 C water and Cu
Qwcu=mc(delta)T=0.05kg(387 J/kgC)(50-20)C+0.25kg(4186 J/kgC)(50-20)C=32000 J
32000J=-mLv=-m(2.26x10^6)
|-m|=14.16g

What am I doing wrong?
 

Answers and Replies

  • #2
Andrew Mason
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Homework Statement


A 50.0g copper calorimeter contains 250 g of water at 20 C. how much steam must be condensed into the water if the final temperature of the system is to reach 50 C?


Homework Equations


Q=-mLv where Q is heat energy, m is mass, Lv is latent vaporization constant of water (3.33x10^5 J/kg).
and
Q=mc(Tf-Ti) where Q is heat energy, m is mass, c is specific heat (steam is 0.48 cal/(gC) or 2010 J/(kgC)), and Tf-Ti is delta T.


The Attempt at a Solution


100 C steam -> 100 C water -> 50 C water and Cu
Qwcu=mc(delta)T=0.05kg(387 J/kgC)(50-20)C+0.25kg(4186 J/kgC)(50-20)C=32000 J
32000J=-mLv=-m(2.26x10^6)
|-m|=14.16g

What am I doing wrong?
You will have to explain what you are doing. You cannot use 50 grams as the mass. As the steam condenses it will add to the mass of the water inside the calorimeter.

AM
 
  • #3
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Forgive me I've had limited access to the internet until today.

So would a better approach be to start with the energy lost as the steam condenses in terms of m, and then use substitution? That wouldn't be viable because that would leave me with 3 unknowns wouldn't it?

I understand why what I tried earlier was wrong, but I don't understand where I'm supposed to be starting with this.
 
  • #4
Andrew Mason
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So would a better approach be to start with the energy lost as the steam condenses in terms of m, and then use substitution? That wouldn't be viable because that would leave me with 3 unknowns wouldn't it?
Why three unknowns?

Let x be the quantity of steam that condenses to result in the temperature of the condensed steam plus copper calorimeter plus 250g of water being 50 C. The heat flow to the water/calorimeter is easily determined. We know the mass and the change of temperature. The heat flow from the steam is unknown only because we do not know the mass of the steam. We know the latent head of vaporization. We know the change in temperature of the steam after it condenses. What can you say about these two heat flows? Write out the equation that relates these two heat flows. You have one equation and one unknown. I think you can work it out.

AM
 
Last edited:
  • #5
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I'll give it a shot that way, but then is the "50g copper calorimeter" unnecessary then? Why doesn't that affect anything?

So, x is the quantity of steam that condenses:

Q(water)=(0.25kg+x)(4186 J/(kgC)(50C-20C)
Q(steam)=-x(2.26x10^6)J/kg
Right? How can I relate that?

Hold the phone! I think I get it. So the energy lost by the steam will be the same amount gained by the water/copper system right? Meaning I can just set -Q(steam)=Q(water) right?

So like this?
Q=mc(delta)T=(0.25kg+ms)(4186 J/kgc)(50-20)C+(0.05kg)(387 J/kgc)(50-20)C Which is the equation for the energy required to raise the water to 50 C
then
Qs=-mLv=-ms(2.26x10^6 J/kg), the energy lost when the vapor condenses.
So -Qs=Q
meaning
ms(2.26x10^6 J/kg)=(0.25kg+ms)(4186 J/kgc)(30)C+(0.05kg)(387 J/kgc)(30)C yes?
Only problem is I got the wrong answer that way.
 
Last edited:
  • #6
Andrew Mason
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Sorry, I overlooked the copper calorimeter (50g = .05 kg). You have to find the heat flow in bringing that from 20 C to 50 C as well as the 250 g of water.

Does your Qsteam is take into account the heat of vaporization and going from 100C to 50C? It does not appear to.

AM
 
  • #7
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I think I figured it out (webassign says I was right both in the problem, and in "practice another version"):
100 C Steam -> 100 C water:
Q=-mLv=-m(540.2cal/g)
100 C Water -> 50 C Water:
Q1=m(1 Cal/gC)(100-50)C
Total Energy lost from condensation:
50m+540.2m=590.2m

20 C water and Cu -> 50 C water and Cu
Q2=250g (1 cal/gC)(50-20)C+50g(0.0924 cal/gC)(50-20)=7638.6 cal

So since Qc=-Qh
7638.6 cal=590.2 cal/g*m
m=12.94g

Look solid?

Oh, and I switched to Calories because it seems like it'd always be easier with water, since water's specific heat is always 1, and I wanted a fresh start anyway.
 
Last edited:
  • #8
Andrew Mason
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That looks right.

AM
 

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