1. Heat and Temperature Change
Physics news
Definitions
Heat and Temperature Change
Examples
Heat and Phase Change
Mechanisms of Heat Flow
Conduction
Convection
Radiation

2. Off-Topic: Physics News
Discovery of Gravitational Waves (2016 Feb 11)
Einstein’s General Theory of Relativity
Gravity bends space and time.
Proven correct for 100 years
Also predicts gravitational waves (confirmed yesterday)

3. Off-Topic: Physics News
LIGO detector
Splits laser beam, travel 2 different paths, recombine
Recombination causes interference pattern (Giancoli 12-6)
Gravity wave passing through both paths change interference pattern
2 detectors, one in Louisiana, one in Washington state

4. Off-Topic: Physics News
Results
Collision of 2 Black holes
Signal detected simultaneously in WA and LA

5. Heat - Definition Energy Transfer by non-mechanical means
Heating water on stove (chemical/electrical)
Burning gasoline in engine (chemical)
Touching a hot/cold object (conduction)
Losing heat through glass window (conduction)
Feeling heat from sun (electromagnetic)
Heating food in microwave (electromagnetic)
Each involves energy transfer by non-mechanical means and raising/lowering of temperature
Heat flows from “hot” to “cold”

6. Heat - Definitions Units of Heat. Calorie (food) Mechanical Equivalent
Calorie - heat required to raise 1 g water 1°C
Kilocalorie calories
Calorie (food)
Actually uses kilocalories
Heat given off by incinerating
Mechanical Equivalent
1 cal = J 1 kcal = kJ
Mechanical equivalent – Friction - > Heat
Electrical equivalent – Electrical Energy - > Heat

7. Heat and Internal Energy
Total thermal energy contained.
𝑈=𝑁 1 2 𝑚 𝑣 2 = 3 2 𝑁𝑘𝑇= 3 2 𝑛𝑅𝑇
Heat
Non-mechanical input or output
∆𝑈=𝑄 U Q

8. Heat and Temperature Change
Adding heat to object causes temperature change.
𝑄=𝑚𝑐∆𝑇
Q – Quantity of Heat (J) (cal)
m – Mass (kg)
c – Specific Heat ( J/kg C°)
ΔT – Temperature change (C°)

9. Specific Heat and Temperature Change
Specific Heat measures ability to absorb heat and distribute to molecules.
𝑄=𝑚𝑐∆𝑇
Table 14-1
Water, wood, etc
Why is water a coolant?

10. Example 14-2 Empty 𝑄=𝑚𝑐∆𝑇=(20 𝑘𝑔)(450 𝐽 𝑘𝑔 𝐶°) 80 𝐶° =720 𝑘𝐽
(a) How much heat is required to raise the temperature of an empty 20 kg vat made of iron from 10°C to 90°C? (b) What if the vat is filled with 20 kg water?
Empty
𝑄=𝑚𝑐∆𝑇=(20 𝑘𝑔)(450 𝐽 𝑘𝑔 𝐶°) 80 𝐶° =720 𝑘𝐽
Filled with 20 kg water
𝑄=𝑚𝑐∆𝑇=(20 𝑘𝑔)(450 𝐽 𝑘𝑔 𝐶°) 80 𝐶°
+ (20 𝑘𝑔)(4186 𝐽 𝑘𝑔 𝐶°) 80 𝐶°
=720 𝑘𝐽+6698 𝑘𝐽=7418 𝑘𝐽

11. Example 14-3 – Hot pan in the sink - 1
5 kg skillet at 200°C placed in sink with 40x50x10 cm water at 20°C. What is final temperature?
Mass of water
40 𝑐𝑚∙50 𝑐𝑚 ∙10 𝑐𝑚=20,000 𝑐𝑚 3 =20 𝐿=20 𝑘𝑔
Requirement for equilibrium
Heat lost by skillet = heat gained by water.
Final temperature same.
Heat exchange equation:
𝑚 𝑠𝑘𝑖𝑙𝑙𝑒𝑡 𝑐 𝑠𝑘𝑖𝑙𝑙𝑒𝑡 ∆ 𝑇 𝑠𝑘𝑖𝑙𝑙𝑒𝑡 = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟
Now just solve..

12. Example 14-3 – Hot pan in the sink - 2
5 kg skillet at 200°C placed in sink with 20 kg water at 20°C. What is final temperature?
Heat exchange equation
𝑚 𝑠𝑘𝑖𝑙𝑙𝑒𝑡 𝑐 𝑠𝑘𝑖𝑙𝑙𝑒𝑡 ∆ 𝑇 𝑠𝑘𝑖𝑙𝑙𝑒𝑡 = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟
Fill in (make each temperature change positive)
5 𝑘𝑔 (450 𝐽 𝑘𝑔 𝐶°) 200 𝐶°− 𝑇 𝑓
= 20 𝑘𝑔 (4186 𝐽 𝑘𝑔 𝐶°) 𝑇 𝑓 −20 𝐶°
450 𝑘𝐽− 2.25 𝑘𝐽 𝐶 𝑇 𝑓 = 83.7 𝑘𝐽 𝐶 𝑇 𝑓 − 𝑘𝐽
𝑘𝐽=(86.0 𝑘𝐽 𝐶 ) 𝑇 𝑓
𝑇 𝑓 =24.7°𝐶
Pan cools more than water heats!

13. Example 14-4 – Cup cools tea.
Heat exchange equation:
Heat lost by tea = heat gained by cup.
𝑚 𝑡𝑒𝑎 𝑐 𝑡𝑒𝑎 ∆ 𝑇 𝑡𝑒𝑎 = 𝑚 𝑐𝑢𝑝 𝑐 𝑐𝑢𝑝 ∆ 𝑇 𝑐𝑢𝑝
Fill in (make temperature changes positive)
0. 2 𝑘𝑔 (4186 𝐽 𝑘𝑔 𝐶°) 95 𝐶°− 𝑇 𝑓
= 0.15 𝑘𝑔 (840 𝐽 𝑘𝑔 𝐶°) 𝑇 𝑓 −25 𝐶°
79,500 𝐽− 837 𝐽 𝐶 𝑇 𝑓 = 126 𝐽 𝐶 𝑇 𝑓 −3150 𝐽
82,650 𝐽=(963 𝐽 𝐶 ) 𝑇 𝑓
𝑇 𝑓 =85.8°𝐶
Again the water (tea) dominates

14. Example 14-5 – 1 hot, 2 cold
Wanted to know specific heat of alloy kg sample heated to 540°C, transferred to 400 g water contained in 200 g aluminum calorimeter cup, all at 10.0°C. Finally temperature observed at 30.5°C.
𝑄 𝑙𝑜𝑠𝑡 𝑎𝑙𝑙𝑜𝑦 = 𝑄 𝑔𝑎𝑖𝑛𝑒𝑑 𝑤𝑎𝑡𝑒𝑟 + 𝑄 𝑔𝑎𝑖𝑛𝑒𝑑 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟
Heat exchange equation: (assume water and cup same ΔT)
𝑚 𝑎𝑙𝑙𝑜𝑦 𝑐 𝑎𝑙𝑙𝑜𝑦 ∆ 𝑇 𝑎𝑙𝑙𝑜𝑦 = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 + 𝑚 𝑐𝑢𝑝 𝑐 𝑐𝑢𝑝 ∆ 𝑇 𝑐𝑢𝑝
0.15 𝑘𝑔 𝑐 𝑎𝑙𝑙𝑜𝑦 °𝐶
= 0.4 𝑘𝑔 (4186 𝐽 𝑘𝑔 𝐶°) 20.5 𝐶°
+ 0.2 𝑘𝑔 (900 𝐽 𝑘𝑔 𝐶°) 20.5 𝐶°
( 𝑘𝑔 𝐶°) 𝑐 𝑎𝑙𝑙𝑜𝑦 =34325 𝐽+3690 𝐽
𝑐 𝑎𝑙𝑙𝑜𝑦 =450 𝐽 𝑘𝑔 𝐶

15. Problem 11 – cold thermometer in hot water
Heat exchange equation:
Heat lost by water = heat gained by thermometer
𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 = 𝑚 𝑔𝑙𝑎𝑠𝑠 𝑐 𝑔𝑙𝑎𝑠𝑠 ∆ 𝑇 𝑔𝑙𝑎𝑠𝑠
Fill in numbers:
0.135 𝑘𝑔 𝐽 𝑘𝑔 𝐶 𝑇 ℎ −39.2 𝐶 = 𝑘𝑔 840 𝐽 𝑘𝑔 𝐶 𝐶
𝐽 𝐶 𝑇 ℎ −22,152 𝐽= 𝐽
𝐽 𝐶 𝑇 ℎ =22,669 𝐽
𝑇 ℎ =40.1 𝐶

16. Problem 12 – hot copper in water + cup
Heat exchange equation: (assume water and cup same ΔT)
Heat lost by copper = heat gained by water + heat gained by cup
𝑚 𝑐𝑢 𝑐 𝑐𝑢 ∆ 𝑇 𝑐𝑢 = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 + 𝑚 𝑎𝑙 𝑐 𝑎𝑙 ∆ 𝑇 𝑎𝑙
Fill in numbers:
0.245 𝑘𝑔 𝐽 𝑘𝑔 𝐶 𝐶− 𝑇 𝑓 = 𝑘𝑔 𝐽 𝑘𝑔 𝐶 𝑇 𝑓 −12 𝐶
𝑘𝑔 900 𝐽 𝑘𝑔 𝐶 𝑇 𝑓 −12 𝐶
27,231 𝐽− 𝐽 𝐶 𝑇 ℎ = 𝐽 𝐶 𝑇 ℎ −41,441 𝐽
𝐽 𝐶 𝑇 ℎ −1566 𝐽
70,239 𝐽= 𝐽 𝐶 𝑇 ℎ
𝑇 ℎ =19.8 𝐶

17. Problem 13 - hot iron in water + vat
Heat exchange equation: (assume water and vat same ΔT)
Heat lost by horseshoe = heat gained by water + heat gained by vat
𝑚 Ω 𝑐 Ω ∆ 𝑇 Ω = 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 + 𝑚 𝑣𝑎𝑡 𝑐 𝑣𝑎𝑡 ∆ 𝑇 𝑣𝑎𝑡
Fill in numbers:
0.40 𝑘𝑔 𝐽 𝑘𝑔 𝐶 𝑇 ℎ −25 𝐶 = 1.35 𝑘𝑔 𝐽 𝑘𝑔 𝐶 5 𝐶
𝑘𝑔 450 𝐽 𝑘𝑔 𝐶 5 𝐶
180 𝐽 𝐶 𝑇 ℎ −4500 𝐽=28,255 𝐽+675 𝐽
180 𝐽 𝐶 𝑇 ℎ =33,430
𝑇 ℎ =185.7 𝐶

18. Problem 15 How long to supply 318,642 J at 750 J/s (watts)
Heat required to heat both coffee and pot
𝑄= 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 + 𝑚 𝑎𝑙 𝑐 𝑎𝑙 ∆ 𝑇 𝑎𝑙
= 0.75 𝑘𝑔 𝐽 𝑘𝑔 𝐶 92 𝐶 𝑘𝑔 900 𝐽 𝑘𝑔 𝐶 92 𝐶
=288,834 𝐽+29,808 𝐽
=318,642 𝐽
How long to supply 318,642 J at 750 J/s (watts)
𝑝𝑜𝑤𝑒𝑟 ∙𝑡𝑖𝑚𝑒=𝑒𝑛𝑒𝑟𝑔𝑦
750 𝐽 𝑠 𝑡=318,742 𝐽
𝑡=425 𝑠 𝑜𝑟 𝑎𝑏𝑜𝑢𝑡 7 𝑚𝑖𝑛𝑢𝑡𝑒𝑠