# Quantity of heat and changes of state

• JulienB
In summary, the conversation involved a discussion of a thermodynamics problem where a mass of water ice was placed in an isolated container with water vapor. After some time, the container was found to contain only liquid water. The conversation revolved around determining the mass of water vapor present in the container. The correct equation to use was determined to be Q1 = -Q2, and the final answer for the mass of water vapor was found to be 30.7 grams. The concept of enthalpy was also briefly discussed.

#### 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: 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.

Julien.

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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

• JulienB
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⋅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?

JulienB said:
@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).

• 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⋅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?

JulienB said:
@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.

• JulienB
@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.

JulienB said:
@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 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.

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.