| Specific and Latent Heats ---------------------------- Joseph F. Alward, PhD Department of Physics University of the Pacific |
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| Heat Unit: One "calorie" (cal) = 4.186 Joules |
Specific Heat Capacity
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Specific heat capacity, c: The amount of heat per gram which must be gained or lost to raise or lower the temperature 1 C. Penny, marble, aluminum are at the same temperature.
Which one stays hot longer after |
Specific Heat Capacity
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DT
= Q/mc ---------------------------------
Same heat is absorbed. |
Specific Heat Capacity
Q = mc
DT
| Substance | Specific Heat (cal/g-C) |
| Water | 1.00 |
| Aluminum | 0.215 |
| Iron | 0.108 |
| Gold | 0.031 |
Specific Heat Capacity
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Which one stays hot longer after being removed? Assume equal masses and heat losses. DT = Q/mc |
|
Why Hot Apple Pies Burn
![]() ccrust / capple < 1 |
Comparing equal masses m, and assuming the same Q loss: DTapple = Q/mcapple
DTcrust =
Q/mccrust |
Hot Water Bottle
| cwater = 1.0
cal/g-C
calcohol = 0.57 cal/g-C
DT = Q/mc |
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The Three Phases of Matter
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Phase Transformations
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Latent Heat of Fusion latent: undisclosed, beneath the surface
|
Fusion = melting or freezing
Question: How much thermal energy in joules
must be absorbed by 50 g of ice at 0 C to
melt it? |
Keeping the Greenhouse Warm
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Placing large vats of water in greenhouses protects fruit from freezing; the heat liberated when the water freezes warms the air. |
Keeping the Wolf Cool
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Panting wolf rids its body of 540 calories of heat energy with each gram of water vapor it exhales. |
Heat of Fusion
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Heat of fusion of gallium = 80 cal/gram Melting point = 29.8 C (85.6 F) |
Heat of Vaporization
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Heat of Vaporization: Water: Lv = 540 cal/g ---------------------------------------
Question: How much heat
Answer: |
Heat Numbers for H20
| Ice to ice:
0.50 cal/g-C Ice to water: 80 cal/g Water to water: 1.0 cal/g-C Water to steam: 540 cal/g Steam to steam: 0.48 cal/g-C |
Temperature versus Heat for Water
![]() 500 grams of -30 C ice is converted to 500 grams of 300 C steam. How many calories are needed? |
Conservation of Energy
| Heat lost by hot object = Heat gained by liquid and cup
|
Example:
100 gram aluminum can with 200 grams of water, all at 20 C. Add 50 gram of copper at 130 C. What is the final temperature? -------------------------------------------------------
Heat gained by water and aluminum |
Conservation of Energy--continued
Heat gained by water |
Heat gained by aluminum
= (100 ) (0.215) (T - 20) = 21.5 T - 430
Heat lost by copper
Heat gained = Heat lost = 598 - 4.6 T 226.1 T = 5028 T = 22.24 C |