Tuesday, February 23, 2021

More than Two Substances and Combining Temperature Change and Phase Changes

 General Chemistry Index

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


––––––––––––––––––––––––––

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 

That's about it!

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?

––––––––––––––––––––––––––

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? 

––––––––––––––––––––––––––

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 


––––––––––––––––––––––––––


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's really not bad.

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).

Friday, February 19, 2021

Thermal Energy and Phase Change

General Chemistry Index

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.


Thermal Energy and Phase Change
Well, now… this…

Why does ice melt? Why does water boil? Or freeze? This must have something to do with heat (thermal energy).

First, think about all those phase changes and all that… Okay, good… moving on…

Indeed, there is something going on with heat! 

Let's think about melting…

There is a certain amount of energy that holds the molecules of a solid in place. If you want to melt the solid, you have to add energy (heat). Doing this will free the molecules.

How much heat? Well…

First off, the amount of heat needed to break the bonds has a name.

Heat of Fusion: the heat absorbed by a unit mass of a given solid at its melting point that completely converts the solid to a liquid at the same temperature: equal to the heat of solidification. (dictionary.com)

(It is not uncommon to just say "heat of fusion" regardless of whether it is melting or freezing)

Now, what does that mean? So, let's go with grams for discussion… for any gram of a thing, there is a certain amount of heat needed to break all the bonds and make it into a liquid. Pretty easy, really.

Each substance has different requirements.  Google will provide lists! 

So, if you understand the concept of Heat of Fusion, then the idea fits directly with boiling (vaporization and condensation). 

Heat of Vaporization: the heat absorbed by a unit of mass of a given liquid at its boiling point that completely converts the liquid to a gas at the same temperature.

(There is a corresponding "heat of condensation" as well that some people might use.) 

So… for any given mass of something, there are certain values for the amount of heat needed to cause a phase change.


So What?

Think about that ice… it heats up gradually until it is all at 0°C. Then, as heat goes into the ice, the temperature does not change. Instead, the molecules become free. Once all of the ice has become water, it starts warming up again.

The same is true at the boiling point.


Well… There must be math, right?

Of course, there is math!

Where m is the mass of the substance and… you're going to love this variable… where ∆Hf is heat of fusion, the heat of the phase change, Q, is found as

Q = m • ∆Hf

So… super easy. Multiply two numbers. 

Same for boiling and condensing. The only difference is that "f" is a "v".

Q = m • ∆Hv

So, mass times some number you look up. How easy can that be?

_________________

Example:

The heat of fusion for brass is 168 J/g. How much heat energy is needed to completely melt 250 grams of brass?

Q = m • ∆Hf 

Q = 250 • 168

Q = 42,000 J

_________________

There has to be some way to make this harder!

Actually, for this in particular, there isn't. However… Of course there's a however! …phase change and temperature change can both be tracked in the same interaction. So, the heat gain, for instance, could be raising the temperature of the ice to the melting point, then melting it, then raising the temperature of the water… 

So, it can be more complex. Not really harder, but definitely longer.

Friday, February 12, 2021

Thermal Energy Change and Transfer

General Chemistry Index

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.


Thermal Energy (Heat) Change and Transfer
Again with the two names? Come on!

Thermal energy… heat… Q… whatever…

In this article we are going with "heat" and "Q" as we talk about the amount of thermal energy that is… doing something. 

So, we discussed in a previous article that the amount of heat related to some temperature change of some substance can be found easily:

Where Q is heat change, where m is mass, is specific heat, and ∆T is change in temperature, 

Q = mc∆T

It's fair to think of it this way…

Q is heat (thermal energy)
m is how much stuff you have measured in mass
is what kind of stuff you have, specifically the physical property of specific heat
∆T is how much did the temperature of the stuff change  

Seems like a pretty good time for an example problem. But first… a little nit-picky thing… that c

c is a constant that relates mass, temperature change and type of substance to heat. So, it has to have units that match what you are working with. Standard science units for heat is joules, and for mass kilograms.

So, for example, c for water is 4128 joules per kilogram per degree C. That means if one kilogram of water cools off by 1°C, it gives off 4128 joules. What's the point? The point is this: depending on what mass and energy units you are using, you need to make sure you are using the correct value for specific heat.
 
Okay… examples…


________________________

EXAMPLE 1—Energy Change:

A cast iron brake rotor of a moderately interesting vehicle, having a mass of 9.5 kg, undergoes an increase in temperature from 25°C to 35°C. How much heat energy was absorbed? The specific heat for cast iron is 449 j/kg • C° .

Step 1: Find the temperature change.

Where Tf represents final temperature and Ti represents initial temperature, then

∆T = Tf - Ti 
∆T = 35 - 25
∆T = 10

Step 2: Use ∆T to find Q

Q = mc∆T
Q = 9.5 • 449 • 10
Q = 42,655 J

________________________


Easy! Now What? There must be some way to make it harder!

Finding something besides Q…

Sometimes, you will be given Q and asked to find one of the other numbers. In such a case, you just use the same equation and solve.

________________________

EXAMPLE 2—Find m (or c or ∆T)

A sample of metal gives off 50,000 J of heat. If the temperature change was 10°C and the specific heat was 250 J/kg • °C, what was the mass?

Given any two of the three things, the algebra is the same, except you solve for a different number.

Q = mc∆T
50,000 = m • 250 • 10
50,000 = m • 2500

50,000 / 2500 = m
20 kg = m


Working with Heat Exchange

The calculations become longer (but equally easy) when dealing with cases where one substance exchanges heat with another substance. Now, there's more parts, but still all easy parts.






Usually, you'll be given a lot of info about one of the substances, then asked to find one part of the info for the other substance. Something like…

A 4 kg chunk of lead with an initial temperature of 300°C is placed into water having an initial temperature of 59.375°C and and cooled off. The final temperature of the mixture is 75°C.

What was the mass of the water?

The specific heats:

water = 4128
lead = 129

You must be joking!

It's not too bad. Because…

HEAT LOST = HEAT GAINED!


So, the lead cooled down, giving off heat. The water heated up, gaining heat.

If we find the heat lost by the lead, that's a number. Then, we use THAT number as the heat gained by the water. Then, we plug in and solve. Actually, it's just a lot of steps, but not hard ones.

Let's do it!

________________________

EXAMPLE 3—Energy Transfer:

A 4 kg chunk of lead with an initial temperature of 300°C is placed into water having an initial temperature of 59.375°C and and cooled off. The final temperature of the mixture is 75°C.

What was the mass of the water?
 
We need to look up the specific heats:

water = 4128
lead = 129

 

PART 1: Heat Lost by the Lead

Step 1: Find the temperature change of the lead.

Where Tf represents final temperature and Ti represents initial temperature, then

∆T = | Tf - T
∆T = 300 - 75
∆T = 225

Step 2: Use ∆T to find Q

Q = mc∆T
Q = 4129225
Q = 116,100 J

Heat Lost = 116.100 J

PART 2: Heat Gained by the Water and finding what is not given

Step 1: Heat Gained

Heat Lost = Heat Gained
116,100 J = Heat Gained


Step 2: Find the temperature change of the water.

Where Tf represents final temperature and Ti represents initial temperature, then

∆T = | Tf - T|
∆T =  | 75 - 59.375 |
∆T = 15.625 


Step 3: Find what is missing (mass, m, in this case)

Q = mc∆T
116,100 = m • 412815.625
116,100 / (4128 • 15.625) = m

calculator… make sure you do the parentheses first!

1.8 kg = m

There was 1.8 kg or water.

________________________

Screen Capture from Video

If you know Q, you can find m, c, or ∆T very easily using simple algebra!

All in all, the whole heat change, heat exchange process is pretty easy. There ARE lots of steps involved, but each step is easy.


Thursday, February 11, 2021

Energy, Heat, Temperature Change, and Thermal Exchange

General Chemistry Index

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.


Energy, Heat, Temperature Change and Thermal Exchange
This doesn't sound too bad, right?

So, energy… heat… temperature change. Seems pretty easy, but…

Since science uses words in very specific ways, we need to do some work to specify what is what.

Energy—the potential to do work. In this and the associated discussions, the general understanding of energy is sort of down-played in favor of the specific type of energy to be explored, which folows.

Heat (also Thermal Energy)—Energy stored with the molecules of a substance; the sum of all the molecular kinetic energy. Usually represented by the letter Q.

Temperature—a number on one of several scales that is an indication of relative average kinetic energy of the molecules being measured.

Temperature change—a quantity that is the difference of two temperatures separated by time. Temperature is represented by the scientific symbol for change, the ∆, and the scientific symbol for temperature, T. Hence, where Tf represents final temperature and Ti represents initial temperature.

∆T = | Tf - T

Hold up! That last thing looked like math! And what… is that the absolute value things! Sheesh! I thought that was just make-believe!


So what does all that mean? Let's say there's this chunk of metal. It has a temperature. That's easy enough. Let's say it's temperature is 20°C. So, that means that the average kinetic energy of all the molecules in that chunk of metal comes to 20°C. 

The sum of all that energy is the thermal energy. You could do something with that energy. Say you put that chunk of metal into a small beaker of cool water… the water would heat up.

Now, if you think about it, the bigger the chunk of metal, the more thermal energy it can store. And the hotter it is to start with… Suppose you had a HUGE chunk of metal at 300°C. If you put that into water, it would cause some of the water to boil! All of the water would heat up!

Intuition would reveal that bigger things have more thermal energy than small things made of the same material. A pound of hot lead will heat up more water that an ounce of hot lead. Further, if something starts off hotter, it has more energy than if it starts off lukewarm.

It's also not a great leap to realize that, as the water heats up—that is to say, gains thermal energy (heat)—the chunk of metal will cool off (lose heat.)

This is heat exchange—the transfer of thermal energy from one thing to another; the higher temperature thing loses heat and the lower temperature thing gains heat.

Now, the law of conservation says that energy cannot be created or destroyed, so… THIS IS HUGE…

Heat loss = Heat gained

And… somewhere up in all those words, we said that the symbol for heat was Q, so…

Qlost = Qgained

or 

Q- = Q+

or

Ql = Qg

Now, we just need to know what Q is in terms of temperature change and how much stuff we have.

Let's go with that intuition again… The hotter to start with… so if it goes from real hot to cool, more energy is given off. This is the temperature change quantity, ∆T.

More stuff has more heat… so… how do we measure "more" or less with regards to matter? Mass, which is the symbol m.

One more thing to consider… Sometimes, things at the same temperature seem to be "hotter" than others. If you have a plastic cup and a glass cup that are both the same temperature (such as if they come out of the hot sink water or dishwasher), the glass cup feels hotter. There must be some factor related to what the material is made of. 

There is.

It's called specific heat. It's just a number, but usually, it is a decimal of some sort. It is represented by the letter c.

Okay, so… how much heat does that chunk of metal have in it? If it gives off some of that heat, we can calculate the amount of heat! Very easy…

More mass = more heat!
Bigger temperature change = more heat!
Then that c thing…

Where Q is heat change, where m is mass, c is specific heat, and ∆T is change in temperature, 

Q = mc∆T

So, now it comes down to math! Really?

It's fair to think of it this way…

Q is heat (thermal energy)
m is how much stuff you have
c is what kind of stuff you have
∆T is how much did the temperature of the stuff change  


The good thing about this is all you do is multiply and divide.

That's enough for now. In other articles, the idea of heat exchange will be looked at more closely and more precisely. And, naturally, there will be examples.