1. Heat Transfer Exam Review? Heat Transfer Mechanisms Conduction
Convection
Radiation

2. Heat Conduction 1 Transfer of molecular motion No net mass transfer
Iron bar – Hot end, Cold end
Handout little papers
Randomly transfer to neighbor
Physically get up and carry to other side
Gas pump handle
Other examples

3. Heat Conduction 2 Relates Heat Flow (Q/t) to ΔT across material
Similar problems:
Stress/strain - relates strain (ΔL/Lo) to stress (F/A) and elastic modulus
Thermal expansion - relates expansion relate ΔL/Lo to temperature and coefficient of thermal expansion
Electrical conduction – relates current (I) to voltage (V) and conductivity/resistivity
Equation involves usual suspects:
Temperature difference
Length
Cross-sectional area
Material property
Power = Q/t = (kA/l)ΔT

4. Heat conduction equation 1
Heat Flow across material by conduction:
𝑃𝑜𝑤𝑒𝑟= 𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇
Power in Watts or J/s
Area A in m2
Length l in m
ΔT in C° or K
Thermal conductivity in 𝐽 𝑠−𝑚−𝐶° 𝑜𝑟 𝑤𝑎𝑡𝑡𝑠 𝑚−𝐶°
Always from hot to cold

5. Heat conduction equation 2
Heat Flow across material by conduction:
𝑃𝑜𝑤𝑒𝑟= 𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇
Power in Watts or J/s
Area A in m2
Length l in m
ΔT in C° or K
C° or K
Thermal conductivity in 𝐽 𝑠−𝑚−𝐶° 𝑜𝑟 𝑤𝑎𝑡𝑡𝑠 𝑚−𝐶°

6. Example 14-10 Heat conduction through glass
𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇
=(0.84 𝐽 𝑠 𝑚 𝐶°) 3 𝑚 𝑚 (15°𝐶−14°𝐶)
=788 𝐽 𝑠=788 𝑤𝑎𝑡𝑡𝑠
Heat conduction through “dead air layer”
=( 𝐽 𝑠 𝑚 𝐶°) 3 𝑚 𝑚 (1 𝐶°)
=2.2 𝐽 𝑠=2.2 𝑤𝑎𝑡𝑡𝑠
“Thermopane” – dry air sandwiched between double glass plate
“R values” at hardware store

7. Problem 37
What must be heat flow across wall to maintain 20 C temperature difference
𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇
=(0.84 𝐽 𝑠 𝑚 𝐶°) 16 𝑚 𝑚 (30° 𝐶 −20° 𝐶)
=2240 𝐽 𝑠=2240 𝑤𝑎𝑡𝑡𝑠
About bulbs

8. Problem 39 – (1)
Heat flow must be same, temperature differences must add.
𝑄 𝑡 𝐶𝑢 = 𝑄 𝑡 𝐴𝑙
𝑘 𝐴 𝑙 ∆𝑇 𝐶𝑢 = 𝑘 𝐴 𝑙 ∆𝑇 𝐴𝑙

9. Problem 39 – (2) Since lengths and cross-sectional areas same.
(380 𝑤𝑎𝑡𝑡𝑠 𝑚 𝐶°) °𝐶−𝑇 = (200 𝑤𝑎𝑡𝑡𝑠 𝑚 𝐶°) 𝑇
95,000−380 𝑇 𝐶°=200 𝑇 𝐶°
580 𝑇 𝐶°=95,000
𝑇=163.8° 𝐶

10. Convection Convection mechanisms Wind blowing in window
Warm air rising in room
Convection oven
Land/sea breeze

11. Electromagnetic Radiation
Transfer by electromagnetic radiation
Sun on a warm day
Hand near a radiator
Campfire
Light bulb
Electromagnetic Radiation
Narrow broadcast radiation (microwave, radio, TV)
Broad thermal radiation (blackbody radiation law)

12. Thermal Radiation – frequency vs temperature
Variation with wavelength/frequency and temperature

13. Thermal Radiation animation

14. Thermal radiation imaging
Boston Marathon Bomber - April 2013

15. Blackbody Radiation Law
Object radiates across spectrum (animation)
Total power radiated
𝑃𝑜𝑤𝑒𝑟= 𝑄 𝑡 =𝑒𝜎𝐴 𝑇 4
T (K)
A (m2)
σ x W/m2T4
e is emissivity ( shiny = 0, black =1)
Good emitter = good absorber

16. Example 14-11 2-way radiated heat given off and absorbed
𝑃𝑜𝑤𝑒𝑟=𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑔𝑖𝑣𝑒𝑛 𝑜𝑓𝑓 −𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑎𝑏𝑜𝑠𝑟𝑏𝑒𝑑
=𝑒𝜎𝐴 𝑇 𝑎𝑡ℎ𝑙𝑒𝑡𝑒 4 −𝑒𝜎𝐴 𝑇 𝑤𝑎𝑙𝑙𝑠 4
= ∙ 10 −8 𝑊 𝑚 2 𝑇 𝑚 𝐾 4 − 288 𝐾 4
=120 𝑊𝑎𝑡𝑡𝑠

17. Example 14-12 (radiation part)
Estimate 0.75 L pot as cube 0.1 m on side
Five 0.1 x 0.1 m sides exposed = area m2
𝑃𝑜𝑤𝑒𝑟=𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑔𝑖𝑣𝑒𝑛 𝑜𝑓𝑓 −𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑎𝑏𝑜𝑠𝑟𝑏𝑒𝑑
=𝑒𝜎𝐴 𝑇 𝑝𝑜𝑡 4 −𝑒𝜎𝐴 𝑇 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 4
=𝑒 5.67∙ 10 −8 𝑊 𝑚 2 𝑇 𝑚 𝐾 4 − 293 𝐾 4
=𝑒∙30 𝑊𝑎𝑡𝑡𝑠
For ceramic pot (e = 0.7) 𝑒∙30 𝑊𝑎𝑡𝑡𝑠=21 watts
For shiny pot (e = 0.1) 𝑒∙30 𝑊𝑎𝑡𝑡𝑠=3 watts

18. Example 14-12 (calorimetry part)
A pot losing a heat Q drops a temperature ΔT:
𝑄=𝑚𝑐∆𝑇
A pot losing a heat Q per second drops a temperature ΔT per second:
𝑄 ∆𝑡 =𝑚𝑐 ∆𝑇 ∆𝑡
∆𝑇 ∆𝑡 = 𝑄 ∆𝑡 𝑚𝑐 = 𝑒 30 𝐽 𝑠 𝑘𝑔 𝐽 𝑘𝑔 𝐶° =𝑒 ∙ 𝐶° 𝑠
The ceramic pot loses C°/s and cools about 12° C in a half hour. (1800 s)
The shiny pot loses C°/s and cools about 2° C in a half hour. (1800 s)

19. Celsius vs. Kelvin Degree sizes same!
Kelvin = Celsius with zero shifted to absolute zero.
𝑇 𝐾 =𝑇 𝐶
May use Celsius when only relative changes important.
Thermal Expansion
Specific and Latent Heat problems
Thermal Conduction
Must use Kelvin when absolute temperature important.
Ideal Gas Law
Kinetic Theory
Thermal Radiation
First and Second Law Thermodynamics
Similar to Gauge vs. Absolute Pressure