So far*.

advertisement
Which one feels hotter/colder?
Based on feeling the 2 objects, it’s relatively easy
to tell which one is colder and which one is
hotter
• That is, we can easily tell that there is a
difference in temperature
• Or is there???
Do they have the same temperature?
can humans really feel temperature?
So far….
• Transfer of energy qualitatively
๏ƒผ Energy is transferred into or out of a system via different
“accounts” (Eth or Eph)
• Now, we need to be able to distinguish how
much energy is transferred when changes take
place in a system
• First, we need to start thinking about how
different substances take in or release energy
Have you ever wondered…
• Why, on a hot summer day, the pool is cooler
than the hot cement?
• Why the ocean is cooler than the hot sand?
• EVEN THOUGH both systems are being
supplied the same amount of energy from the
same energy source (the sun)
• This phenomenon is just one of many
examples that demonstrate certain substances
absorb thermal energy at different rates
What factors would determine the rate
at which energy is taken in or
released?
• Heat flow (q) proportional to mass of sample (m)
• Heat flow (q) proportional to temperature change
(โˆ†T)
๐‘ž∝๐‘š
๐‘ž ∝ โˆ†๐‘‡
๐‘ž ∝ ๐‘š ∝ โˆ†๐‘‡ ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘’๐‘“๐‘œ๐‘Ÿ๐‘’ ๐‘ž = ๐‘šโˆ†๐‘‡
• But what would this indicate?
๏ƒผ Any materials with same mass which is supplied w/ same amount
of energy will absorb and release that energy at the same rate
๏ƒผ Our own experience tells us this just isn’t true (pool & cement
example)
๏ƒผ Therefore, something must be missing from the equation
Specific Heat (c)
•
•
•
•
๐‘ž = ๐‘š๐‘โˆ†๐‘‡
Units of specific heat (c)?
๐‘ž
๐ฝ๐‘œ๐‘ข๐‘™๐‘’๐‘  (๐ฝ)
๐ฝ
๐‘=
=
=
๐‘šโˆ†๐‘‡ ๐‘”๐‘Ÿ๐‘Ž๐‘š ๐‘” ∗ โ„ƒ ๐‘” ∗ โ„ƒ
Specific heat is the amount of energy required to
change the temperature of 1g of a substance by
1 0C
It’s a measure of the relative ease w/ which a
substance may be heated or cooled
Physical property of each pure substance
Liquid water has one
of the highest
specific heat values
of any pure
substance
• What change
would require
the most amount
of energy?
• Solid to liquid
• Liquid to gas
Remember as time increases, energy increases
Specific Heat
• So what does it mean for a substance to have
a high specific heat or low specific heat?
๏ƒผ Substance with high specific heat means it takes more energy
to raise its temperature (water, vegetable oil)
๏ƒผ Substance with low specific heat means it takes less energy to
raise its temperature (iron and many other metals)
How do we calculate amount energy
gained/released during phase change?
• Can we still use our original equation?
๐‘ž = ๐‘š๐‘โˆ†๐‘‡
• It wouldn’t make sense to use this equation during the
plateau range on the heat curve because change in
temp = 0
๏ƒผ This would indicate that there is no energy being added/released for a
system to change states of matter
๏ƒผ We know this isn’t true because even though the temp. doesn’t change,
the energy is still being used for something in the system
• So we must introduce similar equation with subtle
differences if we are going to calculate the amount of
energy gained/lost during phase change
Heat of vaporization and heat of fusion
• Heat of vaporization (Hvap): amount of energy
required to convert a liquid into vapor without
a change in temperature (can use for
condensing as well)
• Heat of fusion (Hfus): amount of energy
required to melt one gram of that substance
(can use for freezing as well)
๐‘ž = ๐‘š๐ป๐‘ฃ๐‘Ž๐‘
๐‘ž = ๐‘š๐ป๐‘“๐‘ข๐‘ 
Remember, earlier we said it would require more energy to change liquid water into
gas than it would to change ice into liquid water. Do the following values for back that
claim up?
Hfus = 334 J/g
Hvap = 2260 J/g
Make sense when you consider pulling apart particles of liquid water until they’re
widely separated in a gas is more difficult than simply giving solid water enough energy
to allow the particles to move freely past one another
Practice
• How much energy is required to raise the
temperature of 475 g of water for a pot of tea
from 14 degrees C to 95 degrees C? Answer in
both J and kJ
Practice
• A student observes a loss of 803 J as 141 g of
aluminum cools from 31.7oC to 25.00C.
Calculate the specific heat of aluminum from
these data.
.85 J/goC
Practice
• Calculate the heat flow when 155 g of sodium
freezes. Express the answer in both J and kJ.
Hfus for sodium = 113 J/g
17,515 J
17.5 kJ
Practice
• To what temperature will 545 g of cobalt be
raised if, beginning at 25.0 oC, it gains 3140 J
of energy? Specific heat of cobalt is .46 J/goC
37.5oC
Practice
• Calculate the energy released as 227 g of
sodium vapor condense
Hvap for sodium = 4.27 kJ/g
969 kJ
Practice
• If you are going to heat some water to boiling
to prepare tea, will it take more time or less
time if you start with hot water than if you
start with cold, or will the times be the same?
Download