Transcript Slide 1

Specific Heat
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• Temperature is the average kinetic energy of the particles
in a substance. The SI unit for temperature is the Kelvin.
a. K = C + 273 (10 C = 283 K)
b. C = K – 273 (10 K = -263 C)
• Thermal Energy – the total of both the kinetic and
potential energy of all the particles in a substance.
Specific Heat
Specific heat
• is the amount of heat energy required to
raise the temperature of 1 g of a substance
by 1 °C
• is different for different substances
• in the SI system has units of J/g C
• in the metric system has units of cal/g C
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Units for Energy
• The SI unit for energy is the Joule. English
units for energy include the unit calories.
• 1000 calories (little c) are in one Calorie
(Big C), which is used to measure the energy
in foods (that the human body can make use
of).
4.186 J = 1 cal
1000 cal = 1 Calorie
Have you ever been super hot at the beach, only to
run down to the water and realize it is so cold? How
can this be?
Why does the land heat up faster and cool down
quicker than the water?
Sand =
Water =
664 J / kg C
4184 J / kg C
Why do materials have different specific heats?
water
metal
• Water molecules form strong bonds with each
other;(hydrogen bonding), therefore it takes
more heat energy to break the molecules apart.
• Metals have weaker bonds (metallic – electron
sea model) and do not need as much energy to
break the metal atoms apart.
Learning Check
A. If the same amount of heat is added to two
different substances, the one with a large specific
heat….
1) has a smaller increase in temperature
2) has a greater increase in temperature
B. When ocean water cools, the surrounding air
1) cools 2) warms 3) stays the same
C. Sand in the desert is hot in the day and cool
at night. Sand must have a
1) high specific heat 2) low specific heat
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Solution
A. For the same amount of heat added, a
substance with a large specific heat
1) has a smaller increase in temperature
B. When ocean water cools, the surrounding air
2) warms
C. Sand in the desert is hot in the day and cool at
night. Sand must have a
2) low specific heat
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Examples of Specific Heats
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Specific Heat Equation
Q = m x C(p) x T
Q = Change in Thermal Energy (Joules or cal)
M = mass of substance (Grams of Kilograms)
C = specific heat of substance
(p) = what phase the material is in
T = change in temperature (Tf – Ti)
(Celsius or Kelvin)
 Calculating Heat with Temperature Increase
How many joules are absorbed by 45.2 g of aluminum if its
temperature rises from 12.5 C to 76.8 C? (The specific
heat of Aluminum is 0.897 J/goC
Given: Mass = 45.2 g
Specific heat (C) for aluminum = 0.897 J/g C
Initial temperature = 12.5 C
Final temperature = 76.8 C
Need: Heat (q) in joules (J)
 Calculating Heat Loss
A 225-g sample of hot tea cools from 74.6 C to 22.4 C.
How much heat, in kilojoules, is lost, assuming that tea
has a specific heat of 4.184 J/goC?
Given: Mass = 225 g
Specific Heat (C) for tea = 4.184 J/g C
Initial temperature = 74.6 C
Final temperature = 22.4 C
Need heat (q) in kilojoules (kJ)
 Calculating Mass Using Specific Heat
Ethanol has a specific heat of 2.46 J/g C. When 655 J
are added to a sample of ethanol, its temperature rises
from 18.2 C to 32.8 C. What is the mass of the ethanol
sample?
Given: Heat (q) = 655 J
Specific Heat (C) for ethanol = 2.46 J/g C
Initial temperature = 18.2 C
Final temperature = 32.8 C
Need: Mass of ethanol sample
Need to do a heating curve of Water Test
Heat is transferred to the particles as they heat up. It’s
what causes the molecules to speed up. This is what we
have been calculating.
q = (mass)(c)(∆T)
But what happens when
we go through a phase
change?
Can we use the same
equation?
Remember the Heating/Cooling Curve
for Water?
What did we learn
from this activity?
As we go through
a phase change,
the temperature
doesn’t change!
Q = m x T x C(p)
If we did used the same equation, then the heat added
would always equal 0, because the temperature doesn’t
change, and that can’t be true!
So what do we do? Find a new equation!
Heat of Fusion –
It’s the amount of
heat absorbed in
order to convert
all of a solid into
its liquid.
Heat of
Vaporization– It’s
the amount of heat
absorbed in order to
convert all of a liquid
into its gas form.
Heat of Fusion Calculation – Energy required to
convert all a solid to liquid at a given temperature.
Q = (heat of fusion) x (mass)
Heat of Vaporization Calculation – Energy
required to convert all of a liquid to gas at a given
temperature.
Q = (heat of vaporization) x (mass)
* (Heat of fusion and vaporization values will come
from a chart)
Example: What is the heat required to melt 25.0
grams of ice? Is this Heat of Fusion or Vaporization?
Q = (heat of fusion) x (mass)
(Heat of fusion of water is 334 J/g)
Q = (334 J/g) x (25.0 g) = 8350 Joules
Example: What is the heat required to evaporate 25.0
grams of water? Is this Heat of Fusion or Vaporization?
Q = (heat of vaporization) x (mass)
(Heat of vaporization of water is 2260 J/g)
Q = (2260 J/g) x (25.0 g) = 56500 Joules
So… if I want the total heat required to melt ice
and turn it to steam what would I need?
1) To melt the ice I need to multiply the heat of
fusion with the mass q =(heat of fusion)x(mass)
2) Then, there is moving the temperature from 0 C
to 100C… for this there is a change in temperature
so we can use… q = (mass)x(∆T)x(c)
3) But wait, that just takes us to 100 C, what about
vaporizing the molecules? Well, then we need
q = (heat of vaporization)x(mass)
Practice Problem: What quantity of heat is
required to melt 500.0 g of ice and then heat the
water to steam at 100 oC?
Heat of fusion of ice = 333 J/g
Specific heat of water = 4.2 J/g•K
Heat of vaporization = 2260 J/g
How much heat is required to melt 500. g of ice
and heat the water to steam at 100 oC?
1.
To melt ice
q = (500. g)(333 J/g) = 1.67 x 105 J
2.
To raise water from 0 oC to 100 oC
q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J
3.
To evaporate water at 100 oC
q = (500. g)(2260 J/g) = 1.13 x 106 J
4.
Total heat energy = 1.51 x 106 J = 1510 kJ