In the previous video, we talked about energy balance and ideal gases.
Today, let’s discuss how to handle solids and liquids when it comes to energy balance.
When we have a substance that has a constant specific volume or density, we call that an
incompressible substance.
Most of the time, we can assume solids and liquids to act as incompressible substances
to solve problems.
Now remember we talked about constant pressure and constant volume specific heats?
Well, when it comes to solids and liquids, both cp and cv can be taken to be identical
so we can represent it with just the letter c.
For incompressible substances, just like for ideal gases, the internal energy is only dependent
on temperature.
To find the difference in internal energy, you can use this equation.
Here, we are integrating the specific heat equation given with respect to temperature
since specific heat varies by temperature.
Most of the time though, you can use the simplified version as long as the specific heat for the
range of temperatures is small.
So for example, let’s look at this table.
Notice how from 250 K to 350 K, the difference is extremely small.
In that case, for most problems, integration isn’t necessary.
The units would be kJ/kg.
Now this is written per unit mass, but you can multiply everything by mass to get this
equation which is the one that gets used the most.
The units would be kJ.
To find the change in enthalpy when it comes to incompressible substances, we can use this
equation.
This is specific heat, this is the change in temperature, this is specific volume and
this is the change in pressure.
Now this equation is actually dependent on whether the substance is a solid or a liquid.
If the substance is a solid, then this part is not important since it becomes a negligible
change, so we end up with this.
Which is also equal to the internal energy change.
Now if the substance is a liquid, then it’s dependent on whether it’s a constant pressure
process or a constant temperature process.
If it’s constant pressure, then delta P is zero, so you end up with the same thing
as the solid substance version.
If it’s a constant temperature process, then delta T is zero, in which case, you end
up with this.
Now let’s go through some examples.
This video is heavily reliant on the previous video since we are pretty much doing the same
thing, so if you need a refresh, please check the description.
Let’s take a look at this problem where we have an iron and we need to figure out
how long it would take for the base plate to warm up to 200 degrees Celsius.
First, we will write down what we know.
It’s a 1000-watt iron but we are told that only 90 percent of the heat generated is transferred
to the plate, meaning heat transfer is only 900 watts.
The base plate has a density of 2770 kg per cubic meter, the specific heat is 875, the
surface area of the base plate is 0.03 square meters and it’s 0.5 cm thick, or 0.005 m.
The initial temperature is 22 degrees Celsius and the final temperature is 200 degrees Celsius.
Now we will start with our energy balance equation.
We are taking the base plate as the system.
Energy in minus the energy out is equal to the change in energy of the system.
So what energy goes in?
That’s heat transfer in but remember that this is happening for a certain time period,
so we need to multiply it by delta T.
Now for energy out, there actually isn’t any, since there is no heat loss.
And that’s equal to the change in the internal energy of the system, which in this case,
is the plate.
So how do we figure out the change in the internal energy of the plate?
For that, we can use this equation.
Now looking at this equation, we have everything except for mass.
So, let’s figure it out.
Mass is equal to density times volume.
How do we figure out the volume of the plate?
Well all we need to do is multiply the surface area by the height.
Let’s plug in our values.
Solving gives us the mass of the plate.
Now we can go back to our equation and start plugin in our values.
Let’s solve for T. So it took about 72 seconds for the plate to heat up to 200 degrees Celsius.
In this question, we have steel rods that are entering an oven and we need to figure
out the rate of heat transfer to the rods.
Let’s write down what we know.
The density of the rods is 7833 kg per cubic meter.
The specific heat is 0.465.
The diameter of the rods is 8 cm, or 0.08m, and the velocity in which they travel through
the oven is 2 m per minute.
The rods enter at 30 degrees Celsius, and leave with a temperature of 700 degrees Celsius.
Let’s start with our energy balance equation.
Energy in minus energy out is equal to the change in energy of the system.
What energy goes in?
Well, that’s heat input.
There is no energy loss, and that’s equal the change in internal energy of the system.
We can figure out the internal energy change of the system using this equation.
We have everything we need except for the mass, so let’s figure that out.
Mass is density times volume.
To figure out the volume, all we need to do is use the formula for the volume of a cylinder.
The question is, what is the height of the cylinder, in this case, the length.
We are told that a rod enters the oven at 2m/min, so for our system, we will look at
a time period of 1 min, meaning the length of the cylinder that travels the oven would
be 2 m.
So that’s the length we will use.
Note that we are given the diameter, so make sure to divide it by 2 to get the radius.
Solving gives us the mass of the rod in a 1-minute period.
Let’s go back to our equation and start plugging in our values.
This tells us the total heat that was transferred.
The question wants us to write our answer in rate form, so we just divide this by a
minute, or 60 seconds, since that’s our time period.
And that is equal to 408 kw.
Let’s take a look at one last example.
In this question, we have brass balls that get quenched in water and we have to figure
out the rate at which heat needs to be removed from the water to keep the water temperature
constant at 50 degrees Celsius.
Let’s write down what we know.
The diameter of the balls is 5 cm or 0.05 m.
The density is 8522 kg per cubic meter and the specific heat is 0.385.
The initial temperature of a ball is 120 degrees Celsius and the final temperature is 74 degrees
Celsius.
Lastly, we are told that 100 balls per minute are cooled.
We will take a single brass ball to be the system and write our energy balance equation.
So in simple terms, the ball will let out a certain amount of energy, in other words,
it will dissipate heat once it enters the water.
If we remove that same amount of heat that’s dissipated from the water, then our water
temperature will remain the same.
So, keeping that in mind, our energy in minus energy out is equal to the change in energy
of the system.
There is no energy input, but there is energy out when the ball is in the water because
it dissipates heat and that is equal to the change in internal energy of the ball.
Don’t forget the negative sign since we subtract energy out.
We can find the internal energy using this equation.
We have everything except for the mass, so let’s figure it out.
Mass is density times volume.
The volume of a sphere can be found using this equation.
Let’s plug in our values, keep in mind that we are given the diameter, so remember to
divide it by 2 to get the radius.
Solving gives us the mass.
Now we can go back to our previous equation and plug in our values.
Now this answer we got is for 1 single ball.
In simple terms, this is the amount of energy released from a single ball when it’s in
the water.
Since we have 100 balls per minute, we need to multiply this value by 100.
So this tells us that 988 kJ of energy must be removed from the water every single minute
to keep the water temperature constant.
That should cover the types of problems you will face when it comes to energy balance
with solids and liquids.
Thanks for watching and best of luck with your studies!