Presentation is loading. Please wait.

Presentation is loading. Please wait.

Specific and Latent Heat

Published byΞœΟ…ΟΟΞ―Ξ½Ξ· ΞšΞΏΟ…Ξ²Ξ­Ξ»Ξ·Ο‚ Modified over 6 years ago

Similar presentations

Presentation on theme: "Specific and Latent Heat"β€” Presentation transcript:

1 Specific and Latent Heat
Definitions Heat and Temperature Change Examples Heat and Phase Change Mechanisms of Heat Flow Conduction Convection Radiation

2 Review - Heat and Temperature Change
Adding heat to object causes temperature change. 𝑄=π‘šπ‘βˆ†π‘‡ Q – Quantity of Heat (J) (cal) m – Mass (kg) c – Specific Heat ( J/kg CΒ°) Ξ”T – Temperature change (CΒ°)

3 Temperature and phase change
Block of ice at -40Β°C Heat as ice to 0Β°C Melt at 0Β°C Heat as water to 100Β°C Vaporize at 100Β°C Heat as vapor to ??Β° C

4 Specific and Latent Heat
Specific Heat involves temperature change 𝑄=π‘šπ‘βˆ†π‘‡ Latent Heat involves phase change 𝑄=π‘šπΏ Latent Heat problems involve: No change in temperature Ice - > liquid, liquid -> gas Depends only on mass and Latent Heat

5 Heat and Phase Change Latent heats are enormous, compared to specific heats! (Be careful with extra 0’s and k-prefixes!)

6 Example 14-7 Step 1 – cool water from 20 C to 0C
𝑄 1 =π‘šπ‘βˆ†π‘‡= 1.5 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 20 𝐢 = π‘˜π½ Step 2 – freeze all water at 0 C 𝑄 2 =π‘šπΏ= 1.5 π‘˜π‘” 333,000 𝐽 π‘˜π‘” (no temperature here) =499.5 π‘˜π½ (careful with 0’s and k-prefixes!) Step 3 – cool frozen water from 0 C to -12 C 𝑄 3 =π‘šπ‘βˆ†π‘‡= 1.5 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 12 𝐢 (now ice) =37.8 𝐾𝐽 Total for whole process 𝑄= 𝑄 1 + 𝑄 2 + 𝑄 3 =125.6 π‘˜π½ π‘˜π½+37.8 π‘˜π½=662.9 π‘˜π½

7 Example 14-8 – Does ice melt? (1)
Requirements Heat lost by water = heat gained by ice Final temperature same Final phase same (except at melting or boiling point) Possible outcomes All ice melts, 𝑇β‰₯0 𝐢 Some ice melts, 𝑇=0 𝐢 Some water freezes, 𝑇=0 𝐢 All water freezes, 𝑇≀0 𝐢 Must β€œexperiment” a little, to see which it is……

8 Example 14-8 – Does ice melt? (2)
Since 0 C is where everything gets complicated, find the heats required to bring everything to 0 C Cool all water down to 0 C 𝑄 1 =π‘šπ‘βˆ†π‘‡= 3 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 20 𝐢 = π‘˜π½ Warm all ice up to 0 C 𝑄 2 =π‘šπ‘βˆ†π‘‡= 0.5 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 10 𝐢 =10.5 π‘˜π½ Melt all ice at 0 C 𝑄 3 =π‘šπΏ= 0.5 π‘˜π‘” 333,000 𝐽/π‘˜π‘” =166.5 π‘˜π½ Conclusion: The ice doesn’t have a snowball’s chance! To coexist with the water as ice, it has to bring the water down to at least 0Β°C. To do that it has to absorb kJ, which is more than enough to warm it to 0Β°C and melt it all.

9 Example 14-8 – Does ice melt? (3)
Now you know final temperature must be > 0Β°C To balance the heat: Bring all the ice up to 0Β°C (10.5 π‘˜π½) Melt all the ice at 0Β°C (166.5 π‘˜π½) Bring all the melted ice up above 0Β°C (???) Bring the tea down not quite to 0Β°C ( π‘˜π½ βˆ’ ???) This will simultaneously satisfy: Heat lost by water= heat gained by ice Final temperature same Final phase same (except at melting or boiling point)

10 Example 14-8 – Does ice melt? (4)
Solution π»π‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘‘ 𝑏𝑦 π‘‘π‘’π‘Ž π‘π‘œπ‘œπ‘™π‘–π‘›π‘” π‘‘π‘œ 𝑇 𝑓 >0 = π»π‘’π‘Žπ‘‘ π‘”π‘Žπ‘–π‘›π‘’π‘‘ 𝑏𝑦 𝑖𝑐𝑒 π‘€π‘Žπ‘Ÿπ‘šπ‘–π‘›π‘” π‘‘π‘œ 0 𝐢 + π»π‘’π‘Žπ‘‘ π‘”π‘Žπ‘–π‘›π‘’π‘‘ 𝑏𝑦 𝑖𝑐𝑒 π‘šπ‘’π‘™π‘‘π‘–π‘›π‘” π‘Žπ‘‘ 0𝐢 + π»π‘’π‘Žπ‘‘ π‘”π‘Žπ‘–π‘›π‘’π‘‘ 𝑏𝑦 π‘šπ‘’π‘™π‘‘π‘’π‘‘ 𝑖𝑐𝑒 π‘€π‘Žπ‘Ÿπ‘šπ‘–π‘›π‘” π‘‘π‘œ 𝑇 𝑓 >0 Plug in numbers 3 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 20 πΆβˆ’ 𝑇 𝑓 =10.5 π‘˜π½ π‘˜π½ π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 𝑇 𝑓 βˆ’0 𝐢 π‘˜π½βˆ’( 𝐽 𝐢) 𝑇 𝑓 =10.5 π‘˜π½ π‘˜π½+( 𝐽 𝐢) 𝑇 𝑓 ( 𝐽 𝐢) 𝑇 𝑓 =74.16 𝐽 𝑻 𝒇 =πŸ“.𝟏°π‘ͺ

11 Example 14-8 – Does ice melt? (modified 1)
Double amount of ice to 1 kg Cool all water to 0 C (same) 𝑄 1 =π‘šπ‘βˆ†π‘‡= 3 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 20 𝐢 = π‘˜π½ Warm all ice to 0 C (double) 𝑄 2 =π‘šπ‘βˆ†π‘‡= 1 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 10 𝐢 =21 π‘˜π½ Melt all ice at 0 C (double) 𝑄 3 =π‘šπΏ= 1 π‘˜π‘” 333,000 𝐽/π‘˜π‘” =333 π‘˜π½ Conclusion: The ice definitely comes up to 0Β° C, as that’s the first place the kJ from the water will go. But it doesn’t all melt, as that would consume more energy than comes out of the water!

12 Example 14-8 – Does ice melt? (modified 2)
Now you know final temperature must be = 0Β°C To balance the heat: Bring all ice up to 0Β°C (21 π‘˜π½) Melt some ice at 0Β°C (??) Bring all tea down to 0Β°C ( π‘˜π½) This will simultaneously satisfy: Heat lost by water= heat gained by ice Final temperature same Final phase same (except at melting or boiling point )

13 Example 14-8 – Does ice melt? (modified 3)
Solution π»π‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘‘ 𝑏𝑦 π‘‘π‘’π‘Ž π‘π‘œπ‘œπ‘™π‘–π‘›π‘” π‘‘π‘œ 0𝐢 = π»π‘’π‘Žπ‘‘ π‘”π‘Žπ‘–π‘›π‘’π‘‘ 𝑏𝑦 𝑖𝑐𝑒 π‘€π‘Žπ‘Ÿπ‘šπ‘–π‘›π‘” π‘‘π‘œ 0 𝐢 + π»π‘’π‘Žπ‘‘ π‘”π‘Žπ‘–π‘›π‘’π‘‘ 𝑏𝑦 π‘šπ‘’π‘™π‘‘π‘–π‘›π‘” π‘π‘Žπ‘Ÿπ‘‘ π‘œπ‘“ 𝑖𝑐𝑒 π‘Žπ‘‘ 0𝐢 Plug in numbers π‘˜π½=π‘š 333 π‘˜π½ π‘˜π‘” +21 π‘˜π½ π‘š=0.69 π‘˜π‘” With 1 kg of ice, only 0.69 kg of it melts PS: Another example - McDonald’s coffee with β€œΒ½ inch ice” trick

14 Example 14-9 – Latent heat of mercury
No β€œexperimenting” required – you know final temperature and phase. Solution π»π‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘‘ 𝑏𝑦 π‘€π‘Žπ‘‘π‘’π‘Ÿ π‘π‘œπ‘œπ‘™π‘–π‘›π‘” π‘‘π‘œ 𝐢+β„Žπ‘’π‘Žπ‘‘ π‘™π‘œπ‘ π‘‘ 𝑏𝑦 𝐴𝑙 𝑐𝑒𝑝 π‘π‘œπ‘œπ‘™π‘–π‘›π‘” π‘‘π‘œ 16.5 𝐢 = π»π‘’π‘Žπ‘‘ π‘”π‘Žπ‘–π‘›π‘’π‘‘ 𝑏𝑦 π‘šπ‘’π‘Ÿπ‘π‘’π‘Ÿπ‘¦ 𝑖𝑛 π‘šπ‘’π‘™π‘‘π‘–π‘›π‘” π‘Žπ‘‘ βˆ’39𝐢 + π»π‘’π‘Žπ‘‘ π‘”π‘Žπ‘–π‘›π‘’π‘‘ 𝑏𝑦 π‘šπ‘’π‘™π‘‘π‘’π‘‘ π‘šπ‘’π‘Ÿπ‘π‘’π‘Ÿπ‘¦ π‘€π‘Žπ‘Ÿπ‘šπ‘–π‘›π‘” π‘‘π‘œ 16.5 𝐢 π‘š π‘€π‘Žπ‘‘π‘’π‘Ÿ 𝑐 π‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ† 𝑇 π‘€π‘Žπ‘‘π‘’π‘Ÿ + π‘š 𝑐𝑒𝑝 𝑐 𝑐𝑒𝑝 βˆ† 𝑇 𝑐𝑒𝑝 = π‘š 𝐻𝑔 𝐿 𝐻𝑔 +π‘š 𝐻𝑔 𝑐 𝐻𝑔 βˆ† 𝑇 𝐻𝑔 1.2 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 3.5 𝐢 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 𝐢 = 1 π‘˜π‘” 𝐿 𝐻𝑔 + 1 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 𝐢 17.58 π‘˜π½+1.57 π‘˜π½= 1 π‘˜π‘” 𝐿 𝐻𝑔 π‘˜π½ 𝐿 𝐻𝑔 = π‘˜π½ π‘˜π‘”

15 Problem – 24 - Ice in liquid nitrogen
Don’t need to β€œexperiment” since you know final temperature/phase! Nitrogen already at its boiling point, just need to vaporize. Heat lost by ice = latent heat gained by nitrogen π‘š 𝑖𝑐𝑒 𝑐 𝑖𝑐𝑒 βˆ† 𝑇 𝑖𝑐𝑒 = π‘š π‘›π‘–π‘‘π‘Ÿπ‘œπ‘”π‘’π‘›π‘‘ 𝐿 π‘›π‘–π‘‘π‘Ÿπ‘œπ‘”π‘’π‘› Filling in, using Ξ”C = Ξ”K and 2100 J = 2.1 kJ 0.03 π‘˜π‘” π‘˜π½ π‘˜π‘” 𝐾 πΎβˆ’77 𝐾 = π‘š π‘›π‘–π‘‘π‘Ÿπ‘œπ‘”π‘’π‘› (200 π‘˜π½/π‘˜) π‘˜π½= π‘š π‘›π‘–π‘‘π‘Ÿπ‘œπ‘”π‘’π‘› (200 π‘˜π½/π‘˜) π‘š π‘›π‘–π‘‘π‘œπ‘”π‘’π‘› =0.062 π‘˜π‘”= 62 𝑔

16 Problem – 25 – Ice in Water Don’t need to β€œexperiment” since you know final temperature/phase! Heat lost by water + heat lost by cup = heat gained by ice (-8.5 -> 0) + heat gained by melting ice ( 0 ) + heat gained by melted ice (0 -> 17) π‘š π‘€π‘Žπ‘‘π‘’π‘Ÿ 𝑐 π‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ† 𝑇 π‘€π‘Žπ‘‘π‘’π‘Ÿ + π‘š 𝑐𝑒𝑝 𝑐 𝑐𝑒𝑝 βˆ† 𝑇 𝑐𝑒𝑝 =π‘š 𝑖𝑐𝑒 𝑐 𝑖𝑐𝑒 βˆ† 𝑇 𝑖𝑐𝑒 + π‘š 𝑖𝑐𝑒 𝐿 𝑖𝑐𝑒 + π‘š π‘šπ‘’π‘™π‘‘π‘’π‘‘.𝑖𝑐𝑒 𝑐 π‘šπ‘’π‘™π‘‘π‘’π‘‘.𝑖𝑐𝑒 βˆ† 𝑇 π‘šπ‘’π‘™π‘‘π‘’π‘‘.𝑖𝑐𝑒 0.31 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 3 𝐢 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 3 𝐢 = π‘š 𝑖𝑐𝑒 𝐽 π‘˜π‘” 𝐢 8.5 𝐢 + π‘š 𝑖𝑐𝑒 333,000 𝐽 π‘˜π‘” + π‘š 𝑖𝑐𝑒 𝐽 π‘˜π‘” 𝐢 17 𝐢 3893 𝐽 𝐽= π‘š 𝑖𝑐𝑒 17,850 𝐽 π‘˜π‘” +333,000 𝐽 π‘˜π‘” +71,162 𝐽 π‘˜π‘” π‘š 𝑖𝑐𝑒 = ,012= π‘˜π‘”=9.8 𝑔

17 Problem – 26 – Water in Iron boiler
π‘ƒπ‘œπ‘€π‘’π‘Ÿβˆ™π‘‘π‘–π‘šπ‘’=π»π‘’π‘Žπ‘‘ 𝑖𝑛 To reach boiling point: (52,000 π‘˜π½ β„Ž) π‘‘π‘–π‘šπ‘’ = 830 π‘˜π‘” π‘˜π½ π‘˜π‘” 𝐢 82 𝐢 π‘˜π‘” π‘˜π½ π‘˜π‘” 𝐢 82 𝐢 (52,000 π‘˜π½ β„Ž) π‘‘π‘–π‘šπ‘’ = 284,899 π‘˜π½+8,487 π‘˜π½ π‘‘π‘–π‘šπ‘’=5.64 β„Žπ‘œπ‘’π‘Ÿπ‘  To turn all to steam: (52,000 π‘˜π½ β„Ž) π‘‘π‘–π‘šπ‘’ = 830 π‘˜π‘” π‘˜π½ π‘˜π‘” (52,000 π‘˜π½ β„Ž) π‘‘π‘–π‘šπ‘’ = 1,875,800 π‘˜π½ π‘‘π‘–π‘šπ‘’=36. 07β„Žπ‘œπ‘’π‘Ÿπ‘ 

18 Problem – 28 – Steam and Ice
Don’t need to β€œexperiment” since you know final temperature/phase Heat lost condensing steam (100) + heat lost cooling condensed steam (100 -> 20) = heat gained by melting ice ( 0 ) + heat gained by heating melted ice (0 -> 20) π‘š π‘ π‘‘π‘’π‘Žπ‘š 𝐿 π‘ π‘‘π‘’π‘Žπ‘š +π‘š π‘π‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘‘βˆ’π‘ π‘‘π‘’π‘Žπ‘š 𝑐 π‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ† 𝑇 π‘π‘œπ‘›π‘‘π‘’π‘›π‘ π‘’π‘‘βˆ’π‘ π‘‘π‘’π‘Žπ‘š = π‘š 𝑖𝑐𝑒 𝐿 𝑖𝑐𝑒 + π‘š π‘šπ‘’π‘™π‘‘π‘’π‘‘βˆ’π‘–π‘π‘’ 𝑐 π‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ† 𝑇 π‘šπ‘’π‘™π‘‘π‘’π‘‘βˆ’π‘–π‘π‘’ π‘š π‘ π‘‘π‘’π‘Žπ‘š (2260 π‘˜π½ π‘˜π‘”) + π‘š π‘ π‘‘π‘’π‘Žπ‘š ( π‘˜π½ π‘˜π‘” 𝐢)(80 𝐢) =(1 π‘˜π‘”)(333 π‘˜π½ π‘˜π‘”)+(1 π‘˜π‘”) ( π‘˜π½ π‘˜π‘” 𝐢)(20 𝐢) π‘š π‘ π‘‘π‘’π‘Žπ‘š ( ) π‘˜π½ π‘˜π‘”= π‘˜π½ π‘š π‘ π‘‘π‘’π‘Žπ‘š = =0.16 π‘˜π‘”=160 𝑔

19 Problem – 29 – Mercury heat of fusion
Don’t need to β€œexperiment” since you know final temperature/phase Heat lost by water + heat lost by calorimeter = + heat gained by melting Hg (-39 ) + heat gained by melted Hg (-39 -> 5.06) π‘š π‘€π‘Žπ‘‘π‘’π‘Ÿ 𝑐 π‘€π‘Žπ‘‘π‘’π‘Ÿ βˆ† 𝑇 π‘€π‘Žπ‘‘π‘’π‘Ÿ + π‘š π‘π‘Žπ‘™ 𝑐 π‘π‘Žπ‘™ βˆ† 𝑇 π‘π‘Žπ‘™ = π‘š 𝐻𝑔 𝐿 𝐻𝑔 + π‘š 𝐻𝑔 𝑐 π‘šπ‘’π‘™π‘‘π‘’π‘‘βˆ’π»π‘” βˆ† 𝑇 π‘šπ‘’π‘™π‘‘π‘’π‘‘βˆ’π»π‘” 0.4 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 𝐢 π‘˜π‘” 𝐽 π‘˜π‘” 𝐢 𝐢 =(1 π‘˜π‘”) 𝐿 𝐻𝑔 +(1 π‘˜π‘”)(138 𝐽 π‘˜π‘” 𝐢)(44.06 𝐢) 12,960 𝐽+4,319 𝐽= 1 π‘˜π‘” 𝐿 𝐻𝑔 +6,080 𝐽 𝐿 𝐻𝑔 =11,199 𝐽 π‘˜π‘”

20 Problem – 30 – Bullet penetrating ice
Kinetic energy = heat of melting 1 2 π‘š 𝑏𝑒𝑙𝑙𝑒𝑑 𝑣 2 = π‘š π‘šπ‘’π‘™π‘‘π‘’π‘‘ 𝐿 1 2 (0.07 π‘˜π‘” π‘š 𝑠 2 = π‘š π‘šπ‘’π‘™π‘‘π‘’π‘‘ 333,000 𝐽 π‘˜π‘” 𝐽= π‘š π‘šπ‘’π‘™π‘‘π‘’π‘‘ 333,000 𝐽 π‘˜π‘” π‘š π‘šπ‘’π‘™π‘‘π‘’π‘‘ = π‘˜π‘”=6.6 𝑔

Download ppt "Specific and Latent Heat"

Similar presentations

Liquids and Solids Water.

Liquids and Solids Water.
Energy in Thermal Processes

Energy in Thermal Processes
Specific Heat Capacity and Latent Heat

Specific Heat Capacity and Latent Heat
The rope ladder of a boat hangs over the side of the boat and just touches the water. The ladder rungs are 8 inches apart. How many rungs will be under.

The rope ladder of a boat hangs over the side of the boat and just touches the water. The ladder rungs are 8 inches apart. How many rungs will be under.
Chapter 11 Energy in Thermal Processes Heat and Internal Energy

Chapter 11 Energy in Thermal Processes Heat and Internal Energy
The pressure exerted by evaporated liquid particles on the surface of the liquid is called:

The pressure exerted by evaporated liquid particles on the surface of the liquid is called:
Heat Physics Lecture Notes

Heat Physics Lecture Notes
Chapter 17. Heat is the amount energy transfer due to a temperature difference. All other forms of energy transfer are classified as work. In the picture.

Chapter 17. Heat is the amount energy transfer due to a temperature difference. All other forms of energy transfer are classified as work. In the picture.
THE QUANTITY OF HEAT Β  The thermal energy lost or gained by objects is called heat. One calorie (cal) is the quantity of heat required to change the temperature.

THE QUANTITY OF HEAT Β  The thermal energy lost or gained by objects is called heat. One calorie (cal) is the quantity of heat required to change the temperature.
Lab 12: Heat, Energy, and Temperature This is it!! Today we are going to measure the specific heat of an unknown metal. Important terms: Temperature, T:

Lab 12: Heat, Energy, and Temperature This is it!! Today we are going to measure the specific heat of an unknown metal. Important terms: Temperature, T:
Copyright Β© 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 5 Energy and States of Matter 5.6 Melting and Freezing 5.7 Boiling.

Copyright Β© 2004 Pearson Education Inc., publishing as Benjamin Cummings. 1 Chapter 5 Energy and States of Matter 5.6 Melting and Freezing 5.7 Boiling.
Lecture 3 Heat Chapter opener. When it is cold, warm clothes act as insulators to reduce heat loss from the body to the environment by conduction and convection.

Lecture 3 Heat Chapter opener. When it is cold, warm clothes act as insulators to reduce heat loss from the body to the environment by conduction and convection.
Thermodynamics. Thermodynamics – The study of heat transformation. Temperature – A measure of the average kinetic energy of the particles in an object.average.

Thermodynamics. Thermodynamics – The study of heat transformation. Temperature – A measure of the average kinetic energy of the particles in an object.average.
Thermal Energy. Thermal energy: kinetic energy of the particles of a substance due to their constant, random motion. Heat: Three Methods of Heat Transfer:

Thermal Energy. Thermal energy: kinetic energy of the particles of a substance due to their constant, random motion. Heat: Three Methods of Heat Transfer:
Reaction Energy.

Reaction Energy.
Thermochemical Calculations

Thermochemical Calculations
Energy Solving Quantitative Energy Problems

Energy Solving Quantitative Energy Problems
Thermal Energy and Heat

Thermal Energy and Heat
Heat, Temperature, and Internal Energy

Heat, Temperature, and Internal Energy
Lesson 7: Just A Phase Key Terms. Solid A solid has definite volume and definite shape. The particles in a solid are closely packed and vibrate in relation.

Lesson 7: Just A Phase Key Terms. Solid A solid has definite volume and definite shape. The particles in a solid are closely packed and vibrate in relation.

Similar presentations

Ads by Google