Imagine, in front of you, there is 1 kg of steel on one side and 1 kg of water on the
other.
If we add heat to these 2 objects, in other words, a constant input of energy, we intuitively
know that the steel will heat up faster than the water.
In fact, if we wanted to raise the temperature from 20 degrees to 30 degrees Celsius, it
will take 5 kJ of energy for the steel and 41.8 kJ of energy for the water.
It’s good if we have a property that can compare the ability for energy storage in
different substances.
This is called specific heat.
Formally, we can say that specific heat is the energy required to raise the temperature
of a unit mass of a substance by one degree.
There are 2 different kinds of specific heat that we need to know.
There is specific heat at constant volume and specific heat at constant pressure.
Denoted cv and cp.
So as the names say, in one scenario, it’s the energy required to raise the temperature
of the unit mass of a substance by one degree while we maintain the volume to be constant.
For cp, it’s the same except we hold the pressure constant.
Now to hold the pressure constant, the system has to expand, and that expansion requires
more energy, so you will notice that cp values will always be greater than cv values.
You can find these values on tables like these for all sorts of different substances.
The units for it are these, and you should note that this is equivalent to the Celsius
version, so you can switch them.
Another thing to keep in mind is that cv is related to the internal energy of a system,
where as cp is related to the change in enthalpy of the system.
You might remember that we talked about the ideal gas equation.
Now we are going to relate internal energy to ideal gases.
Interestingly, internal energy for an ideal gas is a function of temperature only.
In simple terms, it’s based only on temperature, regardless of specific volume or pressure.
In fact, enthalpy of an ideal gas is also a function of temperature only.
Now since both internal energy and enthalpy of ideal gases are based only on temperature,
then specific heat is also dependent on temperature only.
There are 2 equations we need to keep in mind with respect to change in internal energy
and enthalpy.
For specific internal energy change, we can use this equation.
The units would be kj/kg.
If you don’t want it per unit mass, then you simply multiply everything by mass, so
we get this.
For the change in enthalpy, we can use this equation.
Notice that for internal energy change, we use cv and for enthalpy change, we use cp.
Another equation that can be used when it comes to specific heat and ideal gases, is
this.
We can use this to figure out cp or cv, as long as we have one of them and the gas constant.
Lastly, there is a property called the specific heat ratio, and it’s simply the ratio you
get when you divide cp by cv.
We talked about polytropic processes before when it comes to moving boundary work.
We used this equation to figure out the boundary work in such a process.
If the system contains ideal gas, then we can use this equation to figure out boundary
work.
Notice that it’s actually just an incorporation of the ideal gas equation.
The units would be kJ.
If you need it per unit mass, then you can find it using this, which would have the units,
kJ/kg.
Now let’s move on to some examples to see how we can use the equations we talked about.
To solve these problems, you should be familiar with the ideal gas equation and energy balance,
so if you need a refresh, please check the description.
Let’s take a look at this question where we have an electrical oil radiator in a room.
After some time, the room heats up, and we need to figure out how long the radiator ran
for, to warm up the room.
Let’s write down what we know.
The volume of the oil is 40 L, or 0.04 cubic meters, and the volume of the room is 50 cubic
meters.
The initial temperature for the room and the oil in the radiator is 10 degrees Celsius.
The radiator has a power rating of 2.4 kW, or 2.4 kJ/s.
There is a heat loss in the room, which is 0.35 kJ/s.
After some time, the temperature of the air is 20 degrees Celsius and the temperature
of the oil in the radiator is 50 degrees Celsius.
We are also given the density and specific heat of the oil.
While it’s not given, we will need the gas constant of air, which we can find from this
table, and that’s the value right here.
Also, keep in mind that the local atmospheric pressure is 100 kPa.
Lastly, we need the specific heat of air at constant volume, and we can find that using
this table.
That’s the value right here.
Wow, that’s a lot of givens.
Now we can write our energy balance equation.
So our energy in minus the energy out is equal to the change of energy in the system.
Now the heater is on for a certain about of time, which is our energy in.
Because it’s on for a certain time period, we need to multiply it by delta T.
Now while the heater is on, we are told that there is a heat loss in the room.
That’s also happening during this process, for a certain amount of time.
That’s our energy out.
Now what changes occur in the system?
We have 2 changes, so that’s the change of internal energy in the air and the change
of internal energy of the oil.
No other changes occur in the system since we have no potential or kinetic energy change.
Now how do we figure out the change in internal energy of the air?
Since we are given the specific heat, and there is a change in internal energy, we can
use this equation.
The same can be applied to the internal energy change of the oil.
So looking at these 2 equations, we are given everything except for the mass of air, and
mass of oil.
First, we will find the mass of air.
To do so, we need to keep in mind that with the parameters given, the air will act as
an ideal gas.
So let’s write down the equation for an ideal gas.
Let’s isolate it for mass.
Now we can plug in the initial values given to us.
Don’t forget, we need our temperature in kelvin to use this equation.
Let’s solve.
Now we need to figure out the mass of the oil, which is much easier.
Mass is density times volume, and we have both of those values.
So let’s plug in our values and solve.
Now that we have the mass, we can find the internal energy change of the air and oil.
We will start with the air.
Let’s plug in our values.
Solving gives us the internal energy change of the air.
Next, we move onto the oil.
So as before, we plug in our values and solve.
Now we go back to our original equation.
Since delta T is common to both terms, we can factor it out like this.
Now we can plug in our values.
Let’s solve and that is the time it took for the temperature of the room to rise from
10 to 50 degrees Celsius.
In this question, we have argon in a piston-cylinder device and it’s compressed in a polytropic
process.
We need to figure out the work produced and the heat transferred during this process.
Let’s write down what we know.
The n value is 1.2, the initial pressure is 120 kPa, and final pressure is 800 kPa.
The initial temperature is 10 degrees Celsius or 283.15 K. Let’s also find the gas constant
of argon and the specific heat at constant volume for argon.
Here are the values we need.
Now this is a closed system because mass doesn’t leave the system but boundary work is definitely
involved.
Keeping that in mind, we have energy in minus the energy out and that’s equal to the energy
change of the system.
Now we don’t know the directions for work and heat transfer, so we are going to assume
the normal convention for heat transfer and work transfer.
The normal convention is to assume that heat is going into the system and work is done
on the system, meaning work output.
And that’s equal to the change in the internal energy of the system since there is no potential
or kinetic energy change.
The internal energy change can be found using this equation.
Since the question wants us to find values in kJ/kg, in other words, per unit mass, we
can eliminate mass from the equation and write it per unit mass like this.
Looks like we have a bunch of things that we need to figure out.
First, we will figure out the initial and final conditions, which include initial specific
volume, final specific volume, and final temperature.
Given the conditions, argon will behave like an ideal gas, so we can use the ideal gas
equation to figure out the initial specific volume.
Let’s isolate the equation.
Now we can plug in our values.
Let’s solve and we get our initial specific volume.
To find the final specific volume, we can use this equation relating the polytropic
process.
Volume is mass multiplied by specific volume.
So, let’s replace it.
Now remember, mass is constant during the whole process since mass doesn’t leave the
system, so we can eliminate it.
Now let’s plug our values in.
Now we can solve for final specific volume.
Next, we need to figure out the final temperature.
We can use the ideal gas equation that compares two states.
Again, volume is mass times specific volume, so let’s replace it.
Going back to what I said, mass is constant, so we can cancel it out.
Let’s plug in our values.
Solving gives us the final temperature.
Now we are going to figure out the boundary work.
For that, we can use the equation to find boundary work when it comes to a polytropic
process and an ideal gas.
Note that the question wants us to figure out the work produced in kJ/kg, which is per
unit mass.
So we are using the specific volume form, meaning mass isn’t required.
Let’s plug in our values.
Since we want our answer in kJ/kg, we need a unity conversion ratio as well.
We get a negative answer, which means this is actually work input.
In other words, we need to input work to compress the gas.
To figure out the heat transferred, we go back to our original equation with the energy
transfer.
Now we can plug in our values.
Make sure to plug in the negative value since we wrote our equation assuming work output.
Let’s solve.
We got a negative answer, so that means it’s heat output, in other words, heat is transferred
out when the gas is compressed.
And those are our answers.