Quantity of heat and changes of state

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Hi everybody! To begin I'd like to say a big thanks to everyone who helped me on the forum lately: I passed my exams recently and without your help I'm not sure I could have done it!

1. Homework Statement


Now I'm on thermodynamics :)
A mass m1 = 150g of water ice of temperature T1 = -10C is put in an isolated container, which already contains a mass m2 of water vapor of temperature T2 = 120C. After a while there is only liquid water in the container at a temperature T3 = 40C. What was the mass m2 of the water vapor?

Fusion heat of ice: Λs = 333kJ/kg
Vaporization heat of water: Λv = 2256 kJ/kg
Specific heat of water ice: cE = 2.06 kJ/(kg⋅K)
Specific heat of water: cW = 4.19 kJ/(kg⋅K)
Specific heat of water vapor: cD = 2.08 kJ/(kg⋅K)

Homework Equations



Q = m⋅c⋅ΔT
Qchange of state = m⋅Λ

The Attempt at a Solution



I think I'm pretty much able to do this, but I do have a question regarding the signs in the equations.

If Q1 and Q2 are the heat needed respectively for m1 and m2 to reach T3, then Q1 = -Q2. If I take the absolute value, that gives me:

m1⋅(10⋅cE + Λs + 40⋅cW) = m2⋅(-20⋅cD + Λv - 60⋅cW)

Is that correct? Another student did the same but took the absolute value of each change of temperature (so had + instead of the two - on the right side of the equation), and we couldn't really find out yet what's right or wrong. Could someone please help us? :)

Btw I come up with m2 ≈ 40g with the minus, and m2 ≈ 30g with the plus. Hard for me to say which is right.

Thanks a lot in advance for your answers.


Julien.
 
Last edited:

Answers and Replies

  • #2
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You have to be very careful with signs when doing heat transfer problems. If Tf < Ti, Q is negative and there has been a net loss of heat. If Tf > Ti, Q is positive and there has been a net gain of heat. Since no heat escapes, we must have:

heat gained + heat lost = 0

Heat gained = - Heat lost

Their should be + on the right side too, so I guess your friend was right
 
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  • #3
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Hi JulianB. Have you learned about the concept of enthalpy in your thermo course yet?
 
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  • #4
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@AbhinavJ @Chestermiller thanks for your answers! Chester I'm not familiar with enthalpy yet. I checked out the wikipedia page, and the formula looks like nothing I've ever seen.

But actually I've now come up with a new answer:

m1⋅(10⋅cE + Λs + 40⋅cW) = -m2⋅(20⋅cD - Λv + 60⋅cW)

And that gives me m2 = 40g. The difference is that I think that m2⋅Λv should be treated negatively, because the change of state goes from gas to water. Is that a correct assumption?
 
  • #5
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@AbhinavJ @Chestermiller thanks for your answers! Chester I'm not familiar with enthalpy yet. I checked out the wikipedia page, and the formula looks like nothing I've ever seen.

But actually I've now come up with a new answer:

m1⋅(10⋅cE + Λs + 40⋅cW) = -m2⋅(20⋅cD - Λv + 60⋅cW)

And that gives me m2 = 40g. The difference is that I think that m2⋅Λv should be treated negatively, because the change of state goes from gas to water. Is that a correct assumption?
No. The right hand side should read +m2⋅(20⋅cD + Λv + 60⋅cW).
 
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  • #6
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@Chestermiller Okay :/ Why is that? Actually I guess I made a mistake in my last post, here is my train of thoughts:

Q1 = -Q2
10⋅m1⋅cE + m1⋅ΛS + 40⋅m1⋅cW = -20⋅m2⋅cD - m1⋅ΛS - 40⋅m1⋅cW

Is the difference between what you wrote and what I wrote because I didn't take the absolute value in my first line?
 
  • #7
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@Chestermiller Okay :/ Why is that? Actually I guess I made a mistake in my last post, here is my train of thoughts:


Q1 = -Q2
10⋅m1⋅cE + m1⋅ΛS +
40⋅m1⋅cW =
-20⋅m2⋅cD - m1⋅ΛS - 40⋅m1⋅cW

Is the difference between what you wrote
and what I wrote because I didn't take the absolute value in my first line?
Let me make it simpler for you. Forget about the formula Q1=-Q2 for a moment. Check the
processes for whether heat is gained or lost, refer to post #2 for this. Put the absolute values of the process in which heat is gained on one side of the '=' sign and for the process in which heat is lost, put its absolute value on the other side of the '=' sign.

When we use the equation Heat gained = - Heat lost, heat lost is always negative as it is just negative heat gained therefore a negative and a negative results in a positive
For example let water at 120C be cooled to 60C. Using mcDeltaT, we get heat gained by water= mS(Tf-Ti) = mS(60-120)= -60mS
which is negative and with the negative sign already there, it will result to be positive. Similar is the case with latent heats.
 
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  • #8
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@AbhinavJ Thank you very much for your answer. I get now a value m2 = m1⋅(10⋅cE + ΛS + 40⋅cW)/(20⋅CD + ΛV + 60⋅cW) = 30.7 g.

Just for curiosity (and to avoid making such mistakes in the future), should I always consider my "small Q's" (I mean by that each change of temperature or of state) positive, or are there any instances where there would be a minus in the equation? Maybe if for some reason I was to treat the case of a mass that gets hotter and then colder?

Julien.
 
  • #9
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10
@AbhinavJ Thank you very much for your answer. I get now a value m2 = m1⋅(10⋅cE + ΛS + 40⋅cW)/(20⋅CD +
ΛV + 60⋅cW) = 30.7 g.

Just for curiosity (and to avoid making such mistakes in the future), should I always
consider my "small Q's" (I mean by that each
change of temperature or of state) positive, or are there any instances where there would be a minus in the equation? Maybe if for some reason I was to treat the case of a
mass that gets hotter and then colder?


Julien.
Yes you can consider each small q as positive given you group all the processes in which heat is lost on one side of '=' and those in which heat is gained on other side. For a mass that gets colder and then hotter consider the net effect(Assuming no phase change occured while heating and cooling) For eg I heat ice at - 20C to - 10C and then i cool it back to - 30C, the net effect is that it cooled to - 30C from - 20C
 
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  • #10
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Super. You guys have been very helpful once again, thanks a lot!


Julien.
 
  • #11
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Here's a different way of thinking about it. Let q represent the heat content per unit mass of liquid water, water vapor, or ice relative to some reference state, such as liquid water at 0 C. So in the initial state, the heat content per unit mass of the ice and the water vapor are:

$$q_{ice,i}=-\Lambda_S-10 c_E$$
$$q_{vapor,i}=100c_W+\Lambda+20 c_D$$

In the final state, the heat content per unit mass of the liquid water is:
$$q_{liquid,f}=40c_W$$

Since the vessel is isolated (insulated), the final heat content of the system must be equal to the initial heat content:
$$m_1q_{ice,i}+m_2q_{vapor,i}=(m_1+m_2)q_{liquid,f}$$
Try this, and see what you get. Doing it this way helps prevent you from messing up on the signs.

Chet
 
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  • #12
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@Chestermiller Nice! I like this method too, it is somewhat a little more intuitive.

Thanks a lot.

Julien.
 

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