Quantity of heat and changes of state

[JulienB]

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: Λ_s = 333kJ/kg
Vaporization heat of water: Λ_v = 2256 kJ/kg
Specific heat of water ice: c_E = 2.06 kJ/(kg⋅K)
Specific heat of water: c_W = 4.19 kJ/(kg⋅K)
Specific heat of water vapor: c_D = 2.08 kJ/(kg⋅K)

Homework Equations

Q = m⋅c⋅ΔT
Q_{change 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⋅c_E + Λ_s + 40⋅c_W) = m₂⋅(-20⋅c_D + Λ_v - 60⋅c_W)

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.

 

[AbhinavJ]

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

 

[Chestermiller]

Hi JulianB. Have you learned about the concept of enthalpy in your thermo course yet?

 

[JulienB]

@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⋅c_E + Λ_s + 40⋅c_W) = -m₂⋅(20⋅c_D - Λ_v + 60⋅c_W)

And that gives me m₂ = 40g. The difference is that I think that m₂⋅Λ_v should be treated negatively, because the change of state goes from gas to water. Is that a correct assumption?

 

[Chestermiller]

@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⋅c_E + Λ_s + 40⋅c_W) = -m₂⋅(20⋅c_D - Λ_v + 60⋅c_W)

And that gives me m₂ = 40g. The difference is that I think that m₂⋅Λ_v 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⋅c_D + Λ_v + 60⋅c_W).

 

[JulienB]

@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⋅c_E + m₁⋅Λ_S + 40⋅m₁⋅c_W = -20⋅m₂⋅c_D - m₁⋅Λ_S - 40⋅m₁⋅c_W

Is the difference between what you wrote and what I wrote because I didn't take the absolute value in my first line?

 

[AbhinavJ]

@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⋅c_E + m₁⋅Λ_S +
40⋅m₁⋅c_W =
-20⋅m₂⋅c_D - m₁⋅Λ_S - 40⋅m₁⋅c_W

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.

 

[JulienB]

@AbhinavJ Thank you very much for your answer. I get now a value m₂ = m₁⋅(10⋅c_E + Λ_S + 40⋅c_W)/(20⋅C_D + Λ_V + 60⋅c_W) = 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.

 

[AbhinavJ]

@AbhinavJ Thank you very much for your answer. I get now a value m₂ = m₁⋅(10⋅c_E + Λ_S + 40⋅c_W)/(20⋅C_D +
Λ_V + 60⋅c_W) = 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 occurred 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

 

[JulienB]

Super. You guys have been very helpful once again, thanks a lot!Julien.

 

[Chestermiller]

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

 

[JulienB]

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

Thanks a lot.

Julien.