# Heat

Heat cannot be created or destroyed.  Heat can only be transferred from one object to another, and heat always flows from warm to cool.

There is no absolute measurement for heat.  The arbitrary unit for measuring heat quantity is the British Thermal Unit, or BTU.  One BTU is defined as the amount of heat required to raise one pound of water one degree Fahrenheit.  (Note:  No time factor is included.)

Suppose one pound of water and one pound of glycerin are suspended above two gas burners.  Each burner will burn exactly the same amount of gas.  The two liquids have thermometers immersed in them.  The burners are lit and each burner transmits the exact same quantity of heat to the liquids.  After some time, the thermometers in the liquids indicate the water is at 150°; the glycerin is at 208°F.  How can this be?  Equal weights of the two liquids were supplied the exact same quantity of heat.  Why does one get warmer than the other?

The answer:  Glycerin has a different capacity for absorbing heat than the water.  This capacity for absorbing heat is called “specific heat.”

Using the definition for a BTU, we can now compare the heat absorbing capacities of various substances to that of water.  Since it takes one BTU to raise one pound of water one degree Fahrenheit, the specific heat of water is given as one.  The specific heat of other substances can now be expressed in comparison to the specific heat (SH) of water, and calculations made to find the quantity of heat needed to be added or removed from a substance to change the temperature of that substance, to heat or cool it.  An equation can now be developed.

BTU = WT x SH x DT

WT = Weight of a substance in pounds

SH = Specific heat of a substance

DT = Temperature change in degrees Fahrenheit

In the example using glycerin, it was found that it takes 0.576 BTU to raise one pound of glycerin one degree Fahrenheit.  Hence, the SH of glycerin is 0.576.

How much heat is required to raise the temperature of 23 lbs. of glycerin 5°F?  Solving the formula: How much heat must be removed to cool 500 lbs. of ice from 32°F to 20°F?  (The SH of ice is 0.48) Many engineering manuals are available with tables showing the specific heat of various substances.  Heatcraft’s refrigeration sizing manual has tables showing the SH of many products, and even shows the SH of some substances above and below their freezing points, since the SH of some substances will differ substantially if frozen or not.  Trane’s air conditioning manual has SH for building materials, some liquids, and some gases.  Many other tables are available, covering every known substance.

In discussing heat, there are two things to keep in mind.  First, temperature is an indication of the intensity of heat.  Second, that BTU is a quantity of heat.  For example, in order to pasteurize milk, the temperature of the milk must be kept between 131°F and 158°F.  The temperature is a measure of the intensity of heat needed to destroy bacteria.  It may have taken 5000 BTU, the quantity of heat, to get the batch of milk to the intensity of heat as indicated by a thermometer.

Enthalpy is a measure of a substance’s heat content and is used for finding the amount of heat needed to accomplish certain processes.

For instance, the quantity of heat required to heat a pound of water from 32°F to 120°F is 88 BTU (actually 87.92 BTU — the slight discrepancy is due to the fact that the SH of water, or any substance, does not have a constant value, but varies slightly with temperature). But, if we heat the same pound of water from 80°F to 120°F, we only need 40 BTU’s.  The pound of water at 80°F already contained 48 BTU’s.

Enthalpy can be added or subtracted to arrive at total heat of a substance.

An example would be cooling air.  When cooling air, as in air conditioning for comfort, total heat needs to be calculated to properly size equipment.  If we are going to cool the air below its dew point, both sensible and latent heat must be removed.  The sum of the sensible and latent heat is the total heat.  In order to find the total heat, we need to know dry bulb temperatures (DB) and the dew point (DP) temperatures.  As an example:

Air at 78°F DB and 60°F DP is to be cooled to 58°F DB and 55°F DP.  Find the total heat to be removed in cooling each pound of air.  (Reference tables are printed showing properties of air and saturated water vapor.)  From a table in the ASHRAE Guide and Data book, Enthalpy of dry air at 78°F DB equals 18.74 BTU.  Enthalpy of vapor in air at 60°F DP is 12.05 BTU.  Total heat content is 18.74 and 12.05 BTU or 30.79 BTU/LB.  This is the total heat per one pound of air at 78°F DB and 60°F DP.

Let’s determine the amount of total heat that must be removed:

 Enthalpy of dry air at 58°F DB is 13.93 BTU Enthalpy of vapor in air at 55°F DP is 10.01 BTU Total heat equals 13.93 + 10.01 or 23.94 BTU 30.79 – 23.94 = 6.85 BTU/LB We need to size the equipment to remove 6.85 BTU’s per pound of air.

It is interesting to note here that all moisture in atmospheric air actually is superheated steam at very low pressure.  In cooling this air and superheated steam, the steam is de-superheated until it reaches a point at which it starts to condense.  This point is called the dew point.

The foregoing information on heat may or may not be used by you in your daily activities.  Hopefully, at least, it will be interesting.

Feature this resource?:
No