Resource Lesson
Heat
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Basic Definitions:
Absolute temperature
(Kelvin) is defined as the random, average translational kinetic energy of the atoms or molecules within a substance. Not, the KE of the object itself. That is, if an ice cube is traveling at 3 m/sec and has a mass of 1 kg, it has a translational KE of ½(1)3
^{2}
= 4.5 J, that KE does not represent its temperature. Its temperature is represented by the statistical average translational KE of an ice molecule within the cube, KE
_{molecule}
= ½mv
_{rms}
^{2}
where v
_{rms}
represents the root mean square velocity of a random molecule within the ice's lattice structure. That velocity would be very small since the temperature of ice is 0ºC or 273 K.
There are three primary
temperature scales
: Kelvin, Centigrade, and Fahrenheit.
Internal energy
is the total energy of all the molecules within a substance; the sum of their average translational kinetic, rotational kinetic and potential energies. All of these energies are measured on the molecular level, not based on what the "gross large scale" object is doing. Compared to an iceberg, a cup of hot water has a much higher temperature but a much smaller total internal energy. An object's internal energy is sometimes refered to as its thermal energy.
Heat
is the transfer of internal energy from one object to another. Heat flows from objects at high temperatures to objects at lower temperatures and ceases when they reach the same temperature.
Heat Transfer Methods:
Three means by which heat is transferred from one location to another are: convection, conduction, radiation
Convection
means that the actual particles circulate from one location to another. This is the type of heat transfer you think about when you say that hot air rises and cold air sinks.
Conduction
is when the heat is transferred through a material by the collisions of adjacent atoms or molecules. Heat always flows from high temperatures to low temperatures. This is analogous to the fact that charged particles always flow from positions of high potential to those of low potential. The formula used to calculate the rate at which heat is conducted through a solid from its warmer side to its cooler side is
k is the coefficient of conductivity is measured in W/Km
ΔQ/t is called the heat current (J/sec or watts)
ΔT is the temperature differential (K)
A is the cross-sectional area (m
^{2}
)
ΔL is the length or thickness of the material (m)
The importance of this formula is that the amount of heat conducted is directly proportional to the temperature gradient and the conductor's cross-sectional area and is inversely proportional to the conductor's length.
For example, if the metal bar shown in the diagram were to be doubled in length, the heat current would be cut in half since ΔL would be doubled. If the temperature of the heat source (in Kelvin) were to be tripled, the heat current would also be tripled since ΔT would be tripled. If the radius of the conductor were to be doubled, the heat current would be quadrupled since the area of a circle equals π r
^{2}
.
Conductivity is also the reason why different materials at "the same temperature" seem to feel warmer or cooler when touched. A wooden table surface, which has a low coefficient of conductivity (0.08 W/Km), seems to feel warmer than an silver tray, which has a high coefficient of conductivity (430 W/Km), resting on its surface. Both are at room temperature, but the metal conducts heat away from your hand more quickly and therefore "feels" cooler.
When the heat current (
ΔQ/t
) has to pass through multiple layers of materials having different coefficients of conductivity, our formula becomes
,
The ratio
L/k
is called the
R value
for each layer and is used as the means of rating insulation.
Let's practice using these conductivity formulas before examining the basic principles and formulas for radiation.
How much heat would flow through an 8-m
^{2}
area of an un-insulated concrete block (k = 0.8 W/Km) house in 24 hours if the temperature is 37ºC on the one side and 22ºC on the other. The block is 25 cm thick.
If the wall were next to be covered on the inside with an 8-m
^{2}
piece of 1.25-cm thick plywood (k = 0.83 W/Km) how much heat would then flow from the one side of the wall to the other in 24 hours?
If a 9-cm thick layer of glass wool insulation (k = 0.04 W/Km) and a 1.0 cm layer of gypsum board (k = 0.17 W/Km) are now added to the wall, how much heat would then flow from the one side of the wall to the other in 24 hours?
Radiation
is when heat is transferred without the need of a medium. This method involves electromagnetic waves. Red-hot radiators are "cooler" than white-hot radiators which are cooler than blue-hot ones. That it, as the wavelength of the emitted electromagnetic radiation decreases, the temperature of the radiator increases.
Before we discuss the mathematics of radiation, let's look at another property of radiators. Dark-colored objects are better radiators and absorbers of radiation, while shiny or light-colored objects are poor absorbers or radiators - but good reflectors. This is because reflection and absorption are contrary processes. The best radiator is called a blackbody since it does not reflect any incident radiation. Whether a substance plays the role of absorber or emitter depends on how its temperature compares to that of the surrounding environment. If its temperature is higher, it radiates energy; if its temperature is lower, it absorbs energy.
The quantity of heat transferred is calculated using Stefan-Boltzmann's Law which is usually stated as one of the following expressions:
ΔQ/Δt = εσAT
^{4}
P = εσAT
^{4}
P/A = εσT
^{4}
ΔQ/Δt is the heat current radiated from the object to its surroundings (J/sec)
P is the power emitted (W) for all wavelengths
P/A is the power emitted per unit area (W/m
^{2}
)
ε is the emissivity (for a perfect blackbody, ε = 1)
σ is Stefan-Boltzmann constant = 5.67 x 10
^{-8}
W/m
^{2}
K
^{4}
A is surface area (for a sphere A = 4π r
^{2}
)
Δt is the time over which the object is radiating energy
T is the radiator's absolute temperature (K)
When the object is placed inside a cavity, then the power (heat current) is expressed as
ΔQ/Δt = εσA(T
_{o}
^{4}
- T
_{c}
^{4}
)
where T
_{o}
is the temperature of the radiating object and T
_{c}
is the temperature of the cavity into which it has been placed. This formula tells us the rate at which heat is exchanged between the radiating object and the cavity. In this situation, the heat current is referred to as either the net power radiated or the net power absorbed by the radiating body.
Notice that when T
_{o}
>>> T
_{c}
[which makes sense when the radiator is at extremely high temperatures and the background (cavity) is at extremely low temperatures] that this equation becomes
ΔQ/Δt = εσAT
_{o}
^{4}
The most important consequence of this law is that hot objects radiate energy at a rate proportional to the fourth power of their absolute Kelvin temperature.
For example, if the radius of a spherical radiator were to be doubled, then the rate at which heat is radiated from the sphere would be quadrupled since surface area equals 4π r
^{2}
. If the temperature (in Kelvin) of the radiant body were to be tripled, the rate would increase by a factor of 3
^{4}
or become 81 times greater.
Now let's take a moment and practice a problem using Stefan-Boltzmann's Law.
A solid metal sphere (emissivity = 0.65) has a radius of 5 cm is heated to 600ºC and suspended from a thin wire in the center of a room whose temperature is 22ºC. At what rate is heat released to the room?
Mechanical Equivalent of Heat:
When the scope of the Law of Conservation of Mechanical Energy is augmented to include changes in thermal energy, it is often referred to as the
1st Law of Thermodynamics
. Within a closed system, this law can be stated as
ΔPE + ΔKE + ΔTE = 0
In mechanical systems, changes in thermal energy are generally caused by the presence of friction. Work done by friction reduces the mechanical energies present (potential and kinetic) and introduces thermal energy, or heat.
Consider a system that consists of a block sliding across a rough horizontal surface which has a coefficient of kinetic friction of μ = 0.2. Previously, we would have just stated that the work done by friction on the block brings the block to a rest. But where did that energy go? It needs to be accounted for. The block's mechanical energy became thermal energy, evidenced by the increase in the temperature of its lower surface. (This effect can easily be shown by rubbing your hands rapidly across each other. As you rub, the inner surfaces of your hands get warmer - KE being transferred to TE.) Although this is still not the entire story - some of the frictional heat would also be absorbed by the table's surface - we are getting closer to reality.
If we now assume that the entire work done by friction can be accounted for by the increase in the temperature of the block, then we can set up the following equation:
W
_{done by friction}
+ ΔTE = 0
fs cos θ = - ΔTE
fs cos 180º = - ΔTE
- fs = - ΔTE
If we let the change in TE represent the heat gained, Q, then ΔTE = mcΔT, where c is the specific heat of the substance measured in J/kg Cº.
(μN)s = mcΔT
(0.2)(mg)s = mcΔT
ΔT = 0.2gs/c
This relationship between loss of mechanical energy and the increase in the thermal energy of a body is called the
mechanical equivalent of heat
. Numerically, it can be stated as 4.186 Joules of mechanical energy equates to 1 calorie of heat.
Refer to the following information for the next three questions.
A 5-kg aluminum block slides from rest down a 1-meter long, 37º incline. When it arrives at the base of the incline it's speed is only 3 m/sec.
How much energy was lost to frictional heat?
What was the coefficient of friction between the block and the incline's surface?
If 880 J of heat are needed to raise the temperature of 1 kg of aluminum by 1 Cº - that is, aluminum has a specific heat of 880 J/kg Cº - how much did the temperature of our block increase after sliding down the incline?
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