heat transfer physics 202 professor lee carkner lecture 14
Post on 18-Dec-2015
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PAL #13 First Law Final temperature of 20 g, 0 C ice cube dropped into 300 g of hot tea
at 90 C. Add up all heats (Q = cm T and Q = Lm) Heat 1: melt ice
Q1 = (333000)(0.02) = 6660 J Heat 2: warm up now melted ice cube
Q2 = (4190)(0.02)(Tf-0) Heat 3: cool down tea
Q3 = (4190)(0.3)(Tf-90) Step 4: add up heat
Q1 + Q2 + Q3 = 0
6660 + 83.8Tf + 1257Tf –113130 = 0
1340.6Tf = 106470
Tf = 79.4 C
Water condenses out of the air onto a cold piece of metal. Due to this condensation, the temperature of the air around the metal,
A) IncreasesB) DecreasesC) Stays the sameD) Fluctuates unpredictablyE) It depends on the temperature of the
metal
Ten joules of heat are added to a cylinder of gas causing the piston at the top to rise. How much work does the piston do?
A) 0 JoulesB) 5 JoulesC) 10 JoulesD) -10 joulesE) You cannot tell from the
information given
Heat Transfer How is heat transferred from one place to
another? What is moving?
In heat transfer the analogous methods are convection and conduction
both a particle and a wave (but not really)
Conduction
The end in the fire experiences a large vibration of the molecules of the metal
This is called conduction The movement of heat from a high
temperature region to a low temperature region through another material
Conductive Heat Transfer The rate at which heat is transferred by
conduction is given byH = Q/t = kA (TH - TC)/L
Where: Q is heat and t is time A is the cross sectional area of the material (in
the direction of heat transfer) T is the temperature (hot or cold)
Thermal Conductivities Metals generally have high k
For Al, k=235 for Cu, k=428 (W/ m K)
For air, k=0.026 for polyurethane foam, k=0.024 Down filled winter coats trap air for insulation
Composite Slabs
H = Q/t = A (TH - TC)/ (L/k) Where (L/k) is the sum of the
ratios of the thickness and thermal conductivity of each layer of the slab
Radiation Energy can be directly transported by photons
The power (in Watts) that is emitted by an
object depends on its temperature (T), its area (A) and it emissivity ()
Pr = AT4
Where is the Stefan-Boltzmann constant = 5.6703 X 10-8 W/m2 K4
T must be in absolute units (Kelvin)
Absorption of Radiation
Pa =ATenv4
Any object thus has a net energy exchange rate with its environment of:
Pn = Pa -Pr = A(Tenv4 - T4)
Blackbody Radiation
They absorb all of the radiation incident on them
Every object whose temperature is above 0 K emits thermal radiation People emit thermal radiation at infrared
wavelengths and thus can be detected at night with IR goggles
Today’s PALa) Consider a house window
with a size of 1 meter by 1.5 meters and 0.75 cm thick. Inside the house it is 20 C and outside it is 10 C. Compare the heat loss through the window by both conduction and radiation. Which is greater?
b) Imagine you replace the window with a double-pane window composed of two 0.25 cm thick windows with 0.25 cm of air in between. What is the heat loss rate due to conduction now?
Convection Hot air (or any fluid) expands and
becomes less dense than the cooler air around it
If the hot air cools as it rises it will eventually fall back down to be re-heated and rise again
Examples: baseboard heating, boiling water, Earth’s atmosphere
Convection Rate Factors
Fluidity
Energy exchange with environment
How rapidly will the material lose heat?
A small temperature difference may result
in not enough density difference to move
Heat Transfer in The Sun
Near the core (where the energy is produced via hydrogen fusion) energy is transported by radiation
About 75% of the way out, the opacity increases to a level where convection becomes dominant
Convection transports the energy to the surface where it radiates away into space
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