The engineering used in energy conservation work is pretty simple. That’s great for a blog like this, because it permits discussion of some thermodynamic topics that can be simpler and more direct than what you’d get in a university-type course.

Specific heat is one such topic.

While we have not yet really considered much about energy, we will need to refer to energy to discuss specific heat, so we’re kind of bootstrapping here. But I think it isn’t too bad.

One point I need to make note of here is that **heat is one manifestation of energy. **The other manifestation of energy is** work**. With that…

Imagine adding heat (aka energy!) to a substance. Like a burner adding heat to a pot of soup. We all know what happens:

The heat enters the soup, and the soup warms up.

Now, this is such a common experience that it may never have occurred to you to think about it. But heat and temperature are two different things.

Heat is what has been added to our soup (besides our ingredients!) But the resulting rise in temperature is a property of the soup itself. The temperature rise is a *result* of adding heat. And specific heat (typically denoted c) is the property that quantifies the relationship between heat addition or subtraction and temperature changes.

We’ll get a little more specific in a minute, but it is important to understand that different materials can have very different specific heat values. For example, the specific heat of water is 1 Btu/lbm-degF, compared with 0.03 Btu/lbm-degF for gold. That’s a huge difference.

Yeah, I know. What’s with the Btu/Lbm-DegF?

As I said, specific heat links energy flow into/out-of a substance with that substance’s temperature. The formulation of this relationship is:

Heat Flow = mass*specific-heat*temperature-change

Or in formula.

dQ=m c dt

Now, let’s see why those units are the way they are. Let us assume we are going to heat 1 gallon of water. This weighs about 8 pounds (By the way, pounds are actually units of force, not mass. The pound-mass is a contrived unit that is 1/32nd of a “slug”)

Let’s also assume that we know we will be heating on a burner that delivers 500 British Thermal Units per hour (Btuh) into the water (there will be some losses to the surrounding atmosphere, the pot, etc., but not too much.)

Let’s also assume that we put the pot on a burner that is already hot, so there is minimal time warming up the burner and pot.

Let’s assume we start with 60 degree water, and, finally:

Let’s assume we let the water heat up for ten minutes.

The first thing we do is figure out the total heat going into the water by converting 10 minutes to 10/60th of an hour and multiplying times our heat rate (or Power) of 500 Btuh (about 150 watts) :

500 Btuh x 10 min x 1 hour/60 min = 83 Btu ==> This equals our “dQ”

We can now take a quick look at our formula:

83 Btu = 8 Lbm * 1 Btu/Lbm-degF * dT

We see that dT is going to be 83/8, or roughly 10 degrees.

Since we started at 60 degrees, our water will be about 70 degrees after ten minutes.

Now consider the same experiment with 8 pounds of gold

83 Btu = 8 Lbm * 0.03 Btu/Lbm-degF * dT.

Here we see that dT is going to be 83/(8*0.03), or roughly 345 degrees!

Since we started at 60 degrees, our gold will be about 400 degrees after ten minutes, quite different from water!

***

Now, I am sure most readers (if there are any) are going to protest here that these numbers are wrong, that you cannot heat up water or gold without some of the heat radiating and/or convecting away. And that is true. But if one went to the trouble of accurately accounting for that lost energy (meaning energy that actually did not wind up in the gold or the water) the formula would not only work, it would work very well at calculating the temperatures.

***

There are a couple of additional pieces of information t0 relay about specific heat.

For one thing, there are two kinds of specific heat. Specific heat at a constant volume (Cv), and specific heat at a constant pressure (Cp).

To understand the difference, imagine a balloon and a jar that both hold 1 gallon of air that are in indoors at 70 degrees.

Imagine it is very cold outside, like zero.

Imagine you take the jar and the balloon outdoors.

Both the balloon and the jar will begin loosing heat in the frigid air. This is our dQ term, and we will assume it is identical for both the balloon and the jar.

Because the balloon can adjust it’s size, the balloon will begin to shrink as it gets colder and colder. In practical terms, what this means is that the pressure of the air inside the balloon is staying the same, even though the volume shrinks in the cold. (more specifically, the balloon is always in pressure equilibrium with the outdoor atmosphere.)

Now consider the jar of air that loses the same amount of heat. Because the jar cannot decrease in size as the air inside it cools, the pressure inside the jar begins to drop, while the volume remains unchanged.

Now it happens to be that Cp is larger than Cv. Because of this the temperature of the air in the balloon will fall more slowly than the jar if the rate of heat extraction is the same. Why?

Recall that we said both objects were losing heat at the same rate, dQ.

dQ = C * dT ==>> dQ / C = dT

Cp is larger than Cv, so

dQ/Cp < dQ/Cv, and

dTp < dTv by substitution

We can speculate why physically the temperature might drop more slowly in the balloon using the following reasoning. When we bring the jar outside, the atmosphere can do no work on the air as it cools. Thus, there is no work energy (heat equivalent) being added to the air in the jar.

In the case of the balloon, pdV work is being performed * on* the balloon as the balloon cools and shrinks, and this work on the ball0on partially offsets the internal energy (and temperature) reduction occurring in the balloon air. Thus the rate of temperature drop is slower. That’s our theory at least, though it is not completely correct.

As a matter fact, the specific heat of air at constant pressure is about 140% of that of the constant volume specific heat in this temperature range (more fun facts, the ratio of Cp to Cv is known as the “specific heat ratio”, usually designated as “k”)

Note as well that the specific heat of a material will change somewhat as the temperature and/or pressure of the material changes. In day to day energy engineering calculations, this is not usually relevant, since we are working in a fairly constrained range of temperatures and pressures (sort of like the way we don’t worry much about relativity when doing velocity calculations here on earth…)

One final thing will be addressed in another post, but I will also mention it here. When a material melts or boils or freezes or condenses (so called phase changes) the heat added or removed from the material **will not result in a temperature change** until the melting, boiling, freezing or condensing activity is complete.

Well, there was quite a bit there ultimately, but the key takeaway is simple:

**The specific heat of a material allows you to understand the temperature change to a material when heat is added to or taken away from it. Provided the material is not actively undergoing a phase change.
**