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Specific and Latent Heat

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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

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