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

Review - 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°)

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

Specific and Latent Heat Specific Heat involves temperature change 𝑄=𝑚𝑐∆𝑇 Latent Heat involves phase change 𝑄=𝑚𝐿 Latent Heat problems involve: No change in temperature Ice - > liquid, liquid -> gas Depends only on mass and Latent Heat

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

Example 14-7 Step 1 – cool water from 20 C to 0C 𝑄 1 =𝑚𝑐∆𝑇= 1.5 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 20 𝐶 =125.58 𝑘𝐽 Step 2 – freeze all water at 0 C 𝑄 2 =𝑚𝐿= 1.5 𝑘𝑔 333,000 𝐽 𝑘𝑔 (no temperature here) =499.5 𝑘𝐽 (careful with 0’s and k-prefixes!) Step 3 – cool frozen water from 0 C to -12 C 𝑄 3 =𝑚𝑐∆𝑇= 1.5 𝑘𝑔 2100 𝐽 𝑘𝑔 𝐶 12 𝐶 (now ice) =37.8 𝐾𝐽 Total for whole process 𝑄= 𝑄 1 + 𝑄 2 + 𝑄 3 =125.6 𝑘𝐽+499.5 𝑘𝐽+37.8 𝑘𝐽=662.9 𝑘𝐽

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, 𝑇≥0 𝐶 Some ice melts, 𝑇=0 𝐶 Some water freezes, 𝑇=0 𝐶 All water freezes, 𝑇≤0 𝐶 Must “experiment” a little, to see which it is……

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 =𝑚𝑐∆𝑇= 3 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 20 𝐶 =251.16 𝑘𝐽 Warm all ice up to 0 C 𝑄 2 =𝑚𝑐∆𝑇= 0.5 𝑘𝑔 2100 𝐽 𝑘𝑔 𝐶 10 𝐶 =10.5 𝑘𝐽 Melt all ice at 0 C 𝑄 3 =𝑚𝐿= 0.5 𝑘𝑔 333,000 𝐽/𝑘𝑔 =166.5 𝑘𝐽 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 0°C. To do that it has to absorb 251.16 kJ, which is more than enough to warm it to 0°C and melt it all.

Example 14-8 – Does ice melt? (3) Now you know final temperature must be > 0°C To balance the heat: Bring all the ice up to 0°C (10.5 𝑘𝐽) Melt all the ice at 0°C (166.5 𝑘𝐽) Bring all the melted ice up above 0°C (???) Bring the tea down not quite to 0°C (251.16 𝑘𝐽 − ???) This will simultaneously satisfy: Heat lost by water= heat gained by ice Final temperature same Final phase same (except at melting or boiling point)

Example 14-8 – Does ice melt? (4) Solution 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑡𝑒𝑎 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑡𝑜 𝑇 𝑓 >0 = 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑖𝑐𝑒 𝑤𝑎𝑟𝑚𝑖𝑛𝑔 𝑡𝑜 0 𝐶 + 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑖𝑐𝑒 𝑚𝑒𝑙𝑡𝑖𝑛𝑔 𝑎𝑡 0𝐶 + 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑚𝑒𝑙𝑡𝑒𝑑 𝑖𝑐𝑒 𝑤𝑎𝑟𝑚𝑖𝑛𝑔 𝑡𝑜 𝑇 𝑓 >0 Plug in numbers 3 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 20 𝐶− 𝑇 𝑓 =10.5 𝑘𝐽+166.5 𝑘𝐽 + 0.5 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 𝑇 𝑓 −0 𝐶 251.16 𝑘𝐽−(12.558 𝐽 𝐶) 𝑇 𝑓 =10.5 𝑘𝐽+166.5 𝑘𝐽+(2.093 𝐽 𝐶) 𝑇 𝑓 (14.651 𝐽 𝐶) 𝑇 𝑓 =74.16 𝐽 𝑻 𝒇 =𝟓.𝟏°𝑪

Example 14-8 – Does ice melt? (modified 1) Double amount of ice to 1 kg Cool all water to 0 C (same) 𝑄 1 =𝑚𝑐∆𝑇= 3 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 20 𝐶 =251.16 𝑘𝐽 Warm all ice to 0 C (double) 𝑄 2 =𝑚𝑐∆𝑇= 1 𝑘𝑔 2100 𝐽 𝑘𝑔 𝐶 10 𝐶 =21 𝑘𝐽 Melt all ice at 0 C (double) 𝑄 3 =𝑚𝐿= 1 𝑘𝑔 333,000 𝐽/𝑘𝑔 =333 𝑘𝐽 Conclusion: The ice definitely comes up to 0° C, as that’s the first place the 251.16 kJ from the water will go. But it doesn’t all melt, as that would consume more energy than comes out of the water!

Example 14-8 – Does ice melt? (modified 2) Now you know final temperature must be = 0°C To balance the heat: Bring all ice up to 0°C (21 𝑘𝐽) Melt some ice at 0°C (??) Bring all tea down to 0°C (251.16 𝑘𝐽) This will simultaneously satisfy: Heat lost by water= heat gained by ice Final temperature same Final phase same (except at melting or boiling point )

Example 14-8 – Does ice melt? (modified 3) Solution 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑡𝑒𝑎 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑡𝑜 0𝐶 = 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑖𝑐𝑒 𝑤𝑎𝑟𝑚𝑖𝑛𝑔 𝑡𝑜 0 𝐶 + 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑚𝑒𝑙𝑡𝑖𝑛𝑔 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑖𝑐𝑒 𝑎𝑡 0𝐶 Plug in numbers 251.16 𝑘𝐽=𝑚 333 𝑘𝐽 𝑘𝑔 +21 𝑘𝐽 𝑚=0.69 𝑘𝑔 With 1 kg of ice, only 0.69 kg of it melts PS: Another example - McDonald’s coffee with “½ inch ice” trick

Example 14-9 – Latent heat of mercury No “experimenting” required – you know final temperature and phase. Solution 𝐻𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑤𝑎𝑡𝑒𝑟 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑡𝑜 16.5 𝐶+ℎ𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 𝐴𝑙 𝑐𝑢𝑝 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝑡𝑜 16.5 𝐶 = 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 𝑖𝑛 𝑚𝑒𝑙𝑡𝑖𝑛𝑔 𝑎𝑡 −39𝐶 + 𝐻𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑚𝑒𝑙𝑡𝑒𝑑 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 𝑤𝑎𝑟𝑚𝑖𝑛𝑔 𝑡𝑜 16.5 𝐶 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 + 𝑚 𝑐𝑢𝑝 𝑐 𝑐𝑢𝑝 ∆ 𝑇 𝑐𝑢𝑝 = 𝑚 𝐻𝑔 𝐿 𝐻𝑔 +𝑚 𝐻𝑔 𝑐 𝐻𝑔 ∆ 𝑇 𝐻𝑔 1.2 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 3.5 𝐶 + 0.5 𝑘𝑔 900 𝐽 𝑘𝑔 𝐶 3.5 𝐶 = 1 𝑘𝑔 𝐿 𝐻𝑔 + 1 𝑘𝑔 140 𝐽 𝑘𝑔 𝐶 55.5 𝐶 17.58 𝑘𝐽+1.57 𝑘𝐽= 1 𝑘𝑔 𝐿 𝐻𝑔 +7.77 𝑘𝐽 𝐿 𝐻𝑔 =11.38 𝑘𝐽 𝑘𝑔

Problem – 24 - Ice in liquid nitrogen Don’t need to “experiment” 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 𝑚 𝑖𝑐𝑒 𝑐 𝑖𝑐𝑒 ∆ 𝑇 𝑖𝑐𝑒 = 𝑚 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛𝑡 𝐿 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛 Filling in, using ΔC = ΔK and 2100 J = 2.1 kJ 0.03 𝑘𝑔 2.100 𝑘𝐽 𝑘𝑔 𝐾 273 𝐾−77 𝐾 = 𝑚 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛 (200 𝑘𝐽/𝑘) 12.348 𝑘𝐽= 𝑚 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛 (200 𝑘𝐽/𝑘) 𝑚 𝑛𝑖𝑡𝑜𝑔𝑒𝑛 =0.062 𝑘𝑔= 62 𝑔

Problem – 25 – Ice in Water Don’t need to “experiment” 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) 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 + 𝑚 𝑐𝑢𝑝 𝑐 𝑐𝑢𝑝 ∆ 𝑇 𝑐𝑢𝑝 =𝑚 𝑖𝑐𝑒 𝑐 𝑖𝑐𝑒 ∆ 𝑇 𝑖𝑐𝑒 + 𝑚 𝑖𝑐𝑒 𝐿 𝑖𝑐𝑒 + 𝑚 𝑚𝑒𝑙𝑡𝑒𝑑.𝑖𝑐𝑒 𝑐 𝑚𝑒𝑙𝑡𝑒𝑑.𝑖𝑐𝑒 ∆ 𝑇 𝑚𝑒𝑙𝑡𝑒𝑑.𝑖𝑐𝑒 0.31 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 3 𝐶 + 0.095 𝑘𝑔 900 𝐽 𝑘𝑔 𝐶 3 𝐶 = 𝑚 𝑖𝑐𝑒 2100 𝐽 𝑘𝑔 𝐶 8.5 𝐶 + 𝑚 𝑖𝑐𝑒 333,000 𝐽 𝑘𝑔 + 𝑚 𝑖𝑐𝑒 4186 𝐽 𝑘𝑔 𝐶 17 𝐶 3893 𝐽+256.5 𝐽= 𝑚 𝑖𝑐𝑒 17,850 𝐽 𝑘𝑔 +333,000 𝐽 𝑘𝑔 +71,162 𝐽 𝑘𝑔 𝑚 𝑖𝑐𝑒 = 4149.5 422,012=0.0098 𝑘𝑔=9.8 𝑔

Problem – 26 – Water in Iron boiler 𝑃𝑜𝑤𝑒𝑟∙𝑡𝑖𝑚𝑒=𝐻𝑒𝑎𝑡 𝑖𝑛 To reach boiling point: (52,000 𝑘𝐽 ℎ) 𝑡𝑖𝑚𝑒 = 830 𝑘𝑔 4.186 𝑘𝐽 𝑘𝑔 𝐶 82 𝐶 + 230 𝑘𝑔 0.45 𝑘𝐽 𝑘𝑔 𝐶 82 𝐶 (52,000 𝑘𝐽 ℎ) 𝑡𝑖𝑚𝑒 = 284,899 𝑘𝐽+8,487 𝑘𝐽 𝑡𝑖𝑚𝑒=5.64 ℎ𝑜𝑢𝑟𝑠 To turn all to steam: (52,000 𝑘𝐽 ℎ) 𝑡𝑖𝑚𝑒 = 830 𝑘𝑔 2260 𝑘𝐽 𝑘𝑔 (52,000 𝑘𝐽 ℎ) 𝑡𝑖𝑚𝑒 = 1,875,800 𝑘𝐽 𝑡𝑖𝑚𝑒=36. 07ℎ𝑜𝑢𝑟𝑠

Problem – 28 – Steam and Ice Don’t need to “experiment” 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) 𝑚 𝑠𝑡𝑒𝑎𝑚 𝐿 𝑠𝑡𝑒𝑎𝑚 +𝑚 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑑−𝑠𝑡𝑒𝑎𝑚 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑒𝑑−𝑠𝑡𝑒𝑎𝑚 = 𝑚 𝑖𝑐𝑒 𝐿 𝑖𝑐𝑒 + 𝑚 𝑚𝑒𝑙𝑡𝑒𝑑−𝑖𝑐𝑒 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑚𝑒𝑙𝑡𝑒𝑑−𝑖𝑐𝑒 𝑚 𝑠𝑡𝑒𝑎𝑚 (2260 𝑘𝐽 𝑘𝑔) + 𝑚 𝑠𝑡𝑒𝑎𝑚 (4.186 𝑘𝐽 𝑘𝑔 𝐶)(80 𝐶) =(1 𝑘𝑔)(333 𝑘𝐽 𝑘𝑔)+(1 𝑘𝑔) (4.186 𝑘𝐽 𝑘𝑔 𝐶)(20 𝐶) 𝑚 𝑠𝑡𝑒𝑎𝑚 (2260+335) 𝑘𝐽 𝑘𝑔= 333+83.7 𝑘𝐽 𝑚 𝑠𝑡𝑒𝑎𝑚 = 416.7 2595 =0.16 𝑘𝑔=160 𝑔

Problem – 29 – Mercury heat of fusion Don’t need to “experiment” 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) 𝑚 𝑤𝑎𝑡𝑒𝑟 𝑐 𝑤𝑎𝑡𝑒𝑟 ∆ 𝑇 𝑤𝑎𝑡𝑒𝑟 + 𝑚 𝑐𝑎𝑙 𝑐 𝑐𝑎𝑙 ∆ 𝑇 𝑐𝑎𝑙 = 𝑚 𝐻𝑔 𝐿 𝐻𝑔 + 𝑚 𝐻𝑔 𝑐 𝑚𝑒𝑙𝑡𝑒𝑑−𝐻𝑔 ∆ 𝑇 𝑚𝑒𝑙𝑡𝑒𝑑−𝐻𝑔 0.4 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶 7.74 𝐶 + 0.62 𝑘𝑔 900 𝐽 𝑘𝑔 𝐶 7.74 𝐶 =(1 𝑘𝑔) 𝐿 𝐻𝑔 +(1 𝑘𝑔)(138 𝐽 𝑘𝑔 𝐶)(44.06 𝐶) 12,960 𝐽+4,319 𝐽= 1 𝑘𝑔 𝐿 𝐻𝑔 +6,080 𝐽 𝐿 𝐻𝑔 =11,199 𝐽 𝑘𝑔

Problem – 30 – Bullet penetrating ice Kinetic energy = heat of melting 1 2 𝑚 𝑏𝑢𝑙𝑙𝑒𝑡 𝑣 2 = 𝑚 𝑚𝑒𝑙𝑡𝑒𝑑 𝐿 1 2 (0.07 𝑘𝑔 250 𝑚 𝑠 2 = 𝑚 𝑚𝑒𝑙𝑡𝑒𝑑 333,000 𝐽 𝑘𝑔 2187.5 𝐽= 𝑚 𝑚𝑒𝑙𝑡𝑒𝑑 333,000 𝐽 𝑘𝑔 𝑚 𝑚𝑒𝑙𝑡𝑒𝑑 =0.0066 𝑘𝑔=6.6 𝑔