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How is Thermal Absorption Measured?
Name: Lynnette
Status: teacher
Grade: other
Location: NY
Country: USA
Date: Fall 2011
Question:
I am teaching architecture students about thermal performance in buildings and - although I have no science background - am coaching them to pay attention to the units of measure as a way of understanding what is at stake. How is the absorption of material quantified in units? I know it is relative to quantity and specific heat capacity but how exactly is it measured?
Replies:
An outline of physical quantities:
Energy
as
- heat contained in a body,
- electricity in storage
- electricity spent
- equivalent of a quantity of natural gas
- sunlight per area per time
units may be
- Joules, (Joules = Watts * seconds),
also expressed as kJ and MJ (kiloJoules and megaJoules)
- kW.hr ( 1kiloWatt-hour= 3600 kJoules)
- BTU's,
- megatons TNT
Pressure of heat energy (more commonly known as "Temperature")
- passive flow of heat energy is always from higher temperature to lower.
so I conceptualize temperature as local _pressure_ of some pervasive diffusing gas-like stuff.
"Stuff" being thermal energy.
"Diffusing" meaning that most flows have some resistance and
most objects have some capacity or inertia,
so that a redistribution between places is never instant, always takes some time.
"Passive" meaning in the absence of heat-pumps
physical mechanisms which interconvert heat energy
to/from mechanical or electrical energy
and transport those between two places.
- units may be degrees F or degrees C
Degrees Kelvin (degK) are the same size as degC,
and so are equivalent for this topic.
Flow of any energy. "Power".
- heat flow through specified area of wall or insulation
- energy delivered as sunlight on a specified area
- energy use rate of electricity
- equivalent use rate of natural gas
- heat transport in a hot-air duct
- useful mechanical power transmitted by a rotating drive shaft.
- waste heat emitted from the radiator of a car.
- units Watts (1 Watt = 1 Joule/second),
also kW, MW (kilowatts, megawatt, gigawatts)
(Electric power plants produce a largish fraction of a giga-watt.)
BTU's/hr, BTU/day, BTU/yr
1 horsepower = ~ 700Watts.
Area-density of energy flow
Joules/sec.meter2
( That's my notation for Joules/(second*meter^2), same as Joules/second/meter2
Feel free to use whichever makes the math feel straighter for you.)
"1 Sun" ~ 1000w/meter2 on sunny desert ground, 1300w/m2 in space around Earth.
Thermal Infrared radiation emitted by objects at a given temperature.
Volume-density of heat energy
Joules / meter^3
Cost in $ of course, given specified prices of electricity, gas, etc.
Mass
Capacity of a specified body (an object, a room, a structure, a ground-area, an atmosphere-volume)
should be energy per unit temperature change. (Joules/ degC).
Specific heat capacity would be capacity per unit mass, such as J/(degC.kg)
This is what usually applies to a specified material.
Personally I think it is important to get the material's mass-density and convert the specific heat
to what I might call "volume-specific heat" . J/(degC.m3)
Because often volume matters more to us than mass in our habitats.
Also I find that the volume-specific heat is more nearly the same for all materials.
Specific heat can be measured a few ways.
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Water is the handy standard for heat capacity, being 4.2 Joule/degC.cm3 and 1cm3 == 1 gram.
You can get a container holding a known mass of water at a known temperature,
get a known mass of the material in question and equilibrate it at a different temperature
then put the material in the water and stir, then measure the final equalized temperature.
The sum of heat energy in the water and material is presumed to be the same throughout this experiment,
so it helps to be in a thermally-insulating enclosure.
E_tot1 = E_tot2
E_tot = E_water + E_material = Ew + Em
E_water = temp_water * mass_water * spec_ht_water = Tw * Mw * SPw
E_material = temp_material * mass_material* spec_ht_material = Tm * Mm * SPm
E_tot = (Tw * Mw) + (Tm * Mm)
There is time1: E_tot1, Tw1, Tm1 and time2: E_tot2,Tw2, Tm2
E_tot1 = (Tw1 * Mw * SPw) + (Tm1 * Mm * SPm)
E_tot2 = (Tw2 * Mw * SPw) + (Tm2 * Mm * SPm)
E_tot1 = E_tot2 ::
(Tw1 * Mw * SPw) + (Tm1 * Mm * SPm) = (Tw2 * Mw * SPw) + (Tm2 * Mm * SPm)
We want to find SPm.
Getting all the SPm's on one side:
(Tm2 * Mm * SPm) - (Tm1 * Mm * SPm) = (Tw1 * Mw * SPw) - (Tw2 * Mw * SPw)
Factoring:
(Tm2-Tm1) * (Mm * SPm) = (Tw1-Tw2) * (Mw * SPw)
Solving for SPm:
SPm = SPw * (Mw/Mm) * ( (Tw1-Tw2) / (Tm2-Tm1) )
Oh, I forgot to say that Tw2 = Tm2 =Tf. The whole mixed bath is supposed to relax to equal temp.
Final answer:
SPm = SPw * (Mw/Mm) * ( (Tw1-Tf) / (Tf-Tm1) )
The above uses degrees C. Who says that thermal Energy is zero at 0 degrees C?
I do not, really, but we can consider the objects' energies relative to those at 0 degrees C,
You can add in a baseline energy term for each object, but they cancel out before vs. after,
so it is valid to have left out the baseline energy of each object as I did above.
It helps if the material is in a bunch of small pieces,
so it does not take long to change the interior temp of the pieces.
Best to use a largish vacuum-insulated coffee bottle
so you can spend a longer time equilibrating without incurring errors due to E_tot changes.
And the water should never be very hot, because steam vapor carries away lots of heat energy.
This kind of measurement might typically be done with something warmed in a 50 degC "convection oven",
then put it in water 0 degC (just poured from big ice-water jug), then the final temp is ~ 20 degC.
It helps if you can use a thermocouple probe that measures 0.1 degC increments.
Then you can skip the ice-water and shoot for a mere 5degC temp rise in the water.
It should be fine to use nails in lieu of "Steel", and brick broken to 0.5-1 inch pieces.
Important to use small not large pieces of glass, plastic or wood,
because those materials conduct heat quite slowly.
Thin plates are as good as small pieces; only the smallest dimension counts much.
But the larger dimensions must also fit in the container and be stirable.
So popsicle sticks might be a decent sample for "Wood".
I kind of wonder if dry-wall crushed to powder will generate a little heat of its own when wetted.
You might need to change from water to oil to measure it perfectly.
The other approach is to invent a really-well-insulated fan-stirred air oven
with repeatable electric power to the heater resistors,
and measure its temperature rise rate with various samples of material placed inside.
Starting from room-temp each time.
No water, larger more intact samples. More inventing and construction.
There must be a side-channel so the air travels in a loop so it passes over all the material.
The fan and the heater are usually in the side-channel,
but the side-channel is inside the same insulation envelope.
Insulation would be at least two layers of 1" polystyrene covered with aluminum foil inside, outside and between.
The structural frame would be outside the insulation, maybe just 2x4's.
You could measure 2'x4' or 4'x8'-foot sheets of particle-board, granite, bunches of 2"x4" beams etc.
The 4'x8' size may be more ambitious because the insulation area is larger, so heat leakage is more problem.
Maybe one of your more math-minded students would figure the heat loss through the insulation in watts for each degreeC of increase.
compare that with the watts from your heater, presumably 200w to 1000w.
There are some <$30 electric power meters out there, maybe even in local hardware stores.
They are digital, and I think they are probably quite accurate.
Jim Swenson
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