## Tuesday, February 23, 2021

### More than Two Substances and Combining Temperature Change and Phase Changes

Where are we going with this? This page will assist in developing the ability to perform calculations involving heat flow, temperature changes, and phase changes by using known values of specific heat, phase change constants, or both.

More than Two Substances and Combining Temperature Change and Phase Changes
So it's going to be like this, eh?

The idea behinds phase change energy is really very intuitive. Stopping to just think about it generally leads to correct ideas and assumptions.

### Combining Temperature Change and Phase Changes

So, you got some ice. It melts. There was some energy that caused that. Then, the melted ice (also known as water) heats up. There was some more energy. The total amount of energy was… the melting energy plus the heating up energy.

Seems like we should be able to say that a lot more mathy, right?

Where

Qgained is the total energy

Qmelt is the energy to melt the ice

Qincrease is the energy to raise the temperature of the water

then…

Qgained = Qmelt + Qincrease

If it feels like I just made up those variable names, it is because I did (except Qgained is used sometimes).

So you just add up all the parts. The total energy is the sum of the energy needed in each step.

Okay… quick review…

To melt something we get this formula:

Qgained = m∆Hf

To freeze something we get this formula:

Qlost = m∆Hf

NOTE that freezing is the same as melting, only it is giving off heat, not taking it in.

To change the temperature of something we get this formula:

Qgained = mc∆T

So, if something melts and heats up… The "real formula" would be…

Qgained = m∆Hf + mc∆T

What if it cools down and freezes? Same, only heat loss, not gained. However, it might be logical to, since the temperature is falling start with the "high energy" piece of the equation, then do the phase change.

So, if something cools down and freezes … The "real formula" would be…

Qlost = mc∆T + m∆Hf

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EXAMPLE 1:

15 grams of ice, beginning at the melting point (0°C) melts and increases in temperature until it is 30°C. How much energy was needed?

Okay… Let see what's given and what we are finding… This is the hardest part. Actually.

What are we supposed to find? Look at the story and find the question mark…

How much energy was needed?

So, find Qgained

What's given?

m = 15 grams

Ti = 0°C

Tf = 30°C

What's an equation that we could use?

Qgained = m∆Hf + mc∆T

So, we need to find c, ∆T, and ∆Hf

We can calculate ∆T

∆T = Tf - Ti

∆T = 30 - 0

∆T = 30

We can look up c and ∆Hf

c = 4.128 J/g • °C

∆H= 334 J/g

Okay, let's relist all the variables that we need… and colorize!

m = 15 grams

∆T = 30

c = 4.128 J/g • °C

∆H= 334 J/g

Plug in and solve…

Qgained = m∆Hf + mc∆T

Qgained = m∆Hf + mc∆T

Qgained = 15 • 334 15 • 4.128 • 30

Qgained = 5010 + 1857.6

Qgained = 6867.6

So… all we have to do is multiply and add?

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Suppose the ice was below freezing when it started?

Well, that's one more piece of the equation… and… one more thing to look up.

So, we have the heat to warm up the ice, the heat to melt the ice and the heat to warm up the water… Three things. Add them up.

Qgained = QincreaseIce + Qmelt + QincreaseWater

PLOT TWIST!!!! Ice has a different specific heat than does water! Don't panic… It's just a different number to multiply.

Also… Come on! This is ridiculous! …the temperature change for the ice and water will be different. Okay, that's not too bad.

GOOD NEWS: The mass is the same for all parts (since it's the same substance).

Thus…

Qgained = mc∆Tice + m∆Hf + mc∆Twater

And, it works the same going from hot to cold, except it would be heat loss.

Qlost = mc∆Twater + m∆Hf + mc∆Tice

How about an example with less narration?

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EXAMPLE 2:

15 grams of ice at a temperature of -10°C heats up and melts, then heats up to a final temperature of 20°C. How much energy was gained?

Values given / looked up:

Ti (ice) = -10°C
Tf (ice) = Ti (water) = 0°C
Tf (water) = 20°C

m = 15 grams

cice = 2.09

cwater = 4.128 J/g • °C

∆H= 334 J/g

∆Tice  =  | 0 - (-10) |
∆Tice  = 10

∆Twater  = | 20 - 0 |
∆Twater  = 20

Let's do each part, then just add…

Formula

Qgained = QincreaseIce + Qmelt + QincreaseWater

Warming the ice…

QincreaseIce  mc∆Tice
QincreaseIce  15 • 2.09 • 10
QincreaseIce  311.5 J

Melting the ice…

Qmelt m∆Hf
Qmelt = 15 • 334
Qmelt = 5010 J

Warming the water…

QincreaseWater = mc∆Twater
QincreaseWater = 15 • 4.128 • 20
QincreaseWater = 1238.4 J

Qgained = QincreaseIce + Qmelt + QincreaseWater
Qgained = 311.5 + 5010 + 1238.4
Qgained = 6559.9

It can all be done in one step, too…

Qgained = mc∆Tice + m∆Hf + mc∆Twater

Qgained = mc∆Tice + m∆Hf + mc∆Twater

Qgained = 15 • 2.09 • 10  +  15 • 334  +  15 • 4.128 • 20

Qgained = 311.5 + 5010 + 1238.4

Qgained = 6559.9

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NOW… Suppose the ice was below freezing when it started, then it melted, heated up boiled and got hotter?
You. Must. Be. Kidding.

So, we have the heat to warm up the ice, the heat to melt the ice and the heat to warm up the water…then the heat to boil the water and then the heat to raise the temperature of the steam. Five things. Add them up.

Qgained = QincreaseIce + Qmelt + QincreaseWater  Qboil + QincreaseSteam

PLOT TWIST!!!! Steam has a different specific heat than ice and water.

Qgained = mc∆Tice + m∆Hf + mc∆Twater  + m∆Hv + mc∆Tsteam

Once more, the hard part is finding all the variables. Looking stuff up… plugging things in the right place… all that…

ALSO!!!!! It does not have to be water. The same principles apply to any substance. You just need to know all the constants:

c for solid

c for liquid

c for gas

∆Hf

∆Hv

This feels like a good time for an example problem!

____________

EXAMPLE 2

A 9 gram piece of ice having an initial temperature of -30°C gains energy, melts, boils, and the steam ends up with a final temperature of 120°C. How much energy was gained?

We need a lot of variables!

First, we can look up some stuff:

c for solid = 2.09

c for liquid = 4.128

c for gas = 1.996

∆Hf = 334

∆Hv = 2257

The mass was given…

m = 9

The ice, water, and steam changed temperatures.

∆Tice = 0 - (-30)

∆Tice = 30

∆Twater = 100 - 0

∆Twater = 100

∆Tsteam = 120 - 100

∆Tsteam = 20

Okay, ready! (Someone should check the math on this one!)

Qgained = QincreaseIce + Qmelt + QincreaseWater  Qboil + QincreaseSteam

Qgained = mc∆T + mHf + mc∆T + mHv + mc∆T

Qgained = (9)(2.09)(30) + (9)(334) + (9)(4.128)(100) + (9)(2257) + (9)(1.996)(20)

Qgained = 564.3 + 3006 + 3715.2 + 20313 + 359.28

Qgained = 27957.78

____________

Okay, so the above covers phase changes. Not bad. Multiplying and adding. It is possible that some would say the hardest part is reading the story and figuring out what all the numbers are. It is possible they would be right!

Dealing with interactions involving more than two substances is equally easy. And the stories are equally complex.

Let's do it!

### Heat Lost and Gained With More Than Two Substances

Oh, this sounds sooooo fun…

Throughout the rest of this article one rule prevails over everything else:

Heat Lost = Heat Gained

This is true regardless of how many different parts of the interaction exist. Hot things give off heat to the cold things and everything ends up at the final temperature and all the heat given off is equal to all the heat gained.

It also leads to a bazillion different possible situations. Five hot things get put into some cool water in which was already seven other things…Yeah… it can be crazy!

BUT!

Heat Lost = Heat Gained

AKA Qlost = Qgained

AND!

Q = mc∆T

So, you can end up with five things with five masses and five initial temperatures going into a (let's use  fancy science words) energy system with seven things having seven different masses and initial temperatures. Also, keep in mind that once (fancy science word) equilibrium is reached EVERYTHING probably has the same final temperature.

So, just to do it, it could look like this…

Suppose thing 1, thing 2 and thing 3 are hot and the are put in to some water that is cool…

Qlost = Qgained

Qthing1 + Qthing2  + Qthing2 = Qwater

now the magic of color…

Qthing1 + Qthing2  + Qthing2 = Qwater

mc∆T + mc∆T + mc∆T = mc∆T

So, you end up with a whole bunch of things to multiply and add. Probably, you'll need to find something, so that will be what you solve for. Let's say you are looking for the "c" for thing 2. Let's say the "story" gives you all the other masses, temperatures, and specific heats… So…

mc∆T + mc∆T + mc∆T = mc∆T

Okay, suppose you plug in and multiply… It's possible you end up with something like this:

200 + 50c + 150 = 450

350 + 50c = 450

subtract 350 from both sides

50c = 100

divide both sides by 50

c = 2

As was said, the variety of situations is huge BUT, it always comes down to this:

Heat Lost = Heat Gained

As long as you know what gave off heat and what gained heat, it's really pretty easy. If you are given initial and final temperatures, you will know, so… that…

Numerous example problems are provided elsewhere (forthcoming).