# Thermal Expansion and Specific Heat

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```					Thermal Expansion and
Specific Heat
16-1 Continued
Thermal Contraction and
Expansion
• Thermal expansion
– An increase in the volume of a material due to
temperature increase

– Particles move farther apart as temperature
increases
• Example
– Glass thermometer
Heat Transfer
• Specific Heat (c)          Specific Heat Values
– amount of energy               (J/(kg·K))
required to raise the   Water             4184
temp. of 1 kg of        Alcohol           2450
material by 1 degree    Aluminum           920
Kelvin                  Carbon (graphite) 710
– units: J/(kg·K)         Sand               664
or J/(g·°C)             Iron               450
Copper             380
Silver             235
Specific heat cont.
• The lower a material’s specific heat, the
more it’s temperature rises when a given
amount of energy is absorbed by a given
mass
Heat Transfer

• Calorimeter
– device used to
measure
changes in
thermal energy

– in an insulated
system,           Coffee cup Calorimeter

heat gained = heat lost
Heat Transfer
Specific Heat Values
• Which sample will take          (J/(kg·K))
longer to heat to        Water              4184
100°C?                   Alcohol            2450
Aluminum            920
Carbon (graphite) 710
Sand                664
Iron                450
50 g Al     50 g Cu      Copper              380
Silver              235

• Al - It has a higher specific heat.
• Al will also take longer to cool down.
Heat Transfer

Q = m ×c × ΔT
Q:    heat (J)
m:    mass (kg)
ΔT:   change in temperature (K or °C)
C:    specific heat (J/kg·K or J/g.oC)

– Q = heat loss
ΔT = Tf - Ti     + Q = heat gain
Heat Transfer
• A 32-g silver spoon cools from 60°C to 20°C.
How much heat is lost by the spoon?

GIVEN:             WORK:
m = 32 g           Q = m·C· ΔT
Ti = 60°C          ΔT = 20°C - 60°C = – 40°C
Tf = 20°C
Q=?                Q = (32g) (.235J/g·°C) (– 40°C)
C = .235 J/g·°C    Q = -300.8 J
Heat Transfer
• How much heat is required to warm 230 g of
water from 12°C to 90°C?

GIVEN:            WORK:
m = 230 g         Q = m·C· ΔT
Ti = 12°C         m = 230 g
Tf = 90°C         ΔT = 90°C - 12°C = 78°C
Q=?               Q = (230 g)(4.184 J/g·oC) (78 C)
o

C= 4.184 J/g·oC   Q = 75,061 J
• A piece of iron at a temperature of 145°C
cools off to 45°C. If the iron has a mass of
10g and a specific heat of 0.449 J/g·oC,
how much heat is given up?
GIVEN:            WORK:
m = 10 g          Q = m·ΔT·C
Ti = 145°C        m = 10 g
Tf = 45°C         ΔT = 45°C - 145°C = -100°C
Q=?               Q = (10 g)(-100oC)(.449 J/g·oC)
C= .449 J/g·oC    Q = - 449 J

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 views: 15 posted: 7/18/2010 language: English pages: 10