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Specific and Latent Heat

Specific and Latent Heat. Definitions Heat and Temperature Change Examples Heat and Phase Change Examples Mechanisms of Heat Flow Conduction Convection Radiation. Heat and Temperature Change . Adding heat to object causes temperature change . Q – Quantity of Heat (J) ( cal )

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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 • Examples • Mechanisms of Heat Flow • Conduction • Convection • Radiation

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

  3. Temperature and phase change • Temperature and phase changes

  4. Specific and Latent Heat • Specific Heat involves temperature change • Latent Heat involves phase change • 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 • Step 2 – freeze all water at 0 C (no temperature here) (careful with 0’s and k-prefixes!) • Step 3 – cool frozen water from 0 C to -12 C (now ice) • Total for whole process

  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, • Some ice melts, • Some water freezes, • All water freezes, Must “experiment” 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 • Warm all ice up to 0 C • Melt all ice at 0 C • 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.

  9. 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 () • Melt all the ice at 0°C () • Bring all the melted ice up above 0°C (???) • Bring the tea down not quite to 0°C () • 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 • Plug in numbers

  11. Example 14-8 – Does ice melt? (modified 1) • Double amount of ice to 1 kg • Cool all water to 0 C (same) • Warm all ice to 0 C (double) • Melt all ice at 0 C (double) • 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 have to 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 = 0°C • To balance the heat: • Bring all ice up to 0°C () • Melt some ice at 0°C (??) • Bring all tea down to 0°C () • 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 • Plug in numbers • With 1 kg of ice, only 0.69 kg of it melts • PS: Another example - McDonald’s coffee with “½ inch ice” trick

  14. Example 14-9 – Latent heat of mercury • No “experimenting” required – you know final temperature and phase. • Solution

  15. 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 • Filling in, using ΔC = ΔK and 2100 J = 2.1 kJ

  16. 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)

  17. Problem – 26 – Water in Iron boiler • To reach boiling point: • To turn all to steam:

  18. 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)

  19. 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)

  20. Problem – 30 – Bullet penetrating ice • Kinetic energy = heat of melting

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