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Transcript
2nd Law of HW 12-2 AND 12-3
Thermodynamics:
• Heat always flows from HOT to COLD
Heat transfer Khan IR Camera
Heat ≠ Temperature
Bozeman temp
• Temperature : proportional to
average kinetic energy of particles
(Units: K (Kelvin) ; oC (Celcius)
• Heat – energy that flows from a
hotter object (higher temperature) to
a colder object (lower temperature)
(Units: J (joule)
WORK HW 12-2
Copy at bottom of p. 128; write
equation on top of 129.
Q = s×m×ΔT
• Q = heat absorbed or lost
+Q if absorbed, -Q if lost
• m = mass in grams
• ΔT = temperature change:
Δ T = Tfinal – Tinitial
• s = specific heat capacity
units of s are J/gºC
NO, values range from 5 to 15
50 ÷ 5 = 10 J
KINETIC ENERGY
#1: 30 ÷ 3 = 10
10
#1: 12 ÷ 3 = 4
4
#1; Higher Average KE
HOT TO COLD
HEAT WILL FLOW FROM HOT (SYSTEM #1)
TO COLD (SYSTEM #2)
#1: 21 ÷ 3 = 7
7
#2: 21 ÷ 3 = 7
7
TEMPERATURE ARE EQUAL (SAME AVERAGE KE; THERMAL =
HEAT EQUILIBRIUM = BALANCE ; TEMPERATURES ARE EQUAL
THERMAL
EQUILIBRIUM
HW 12-3 CONCEPT OF HEAT
Is the amount of material (mass)
important in heat transfer?
B) The cup will have the higher final temperature;
Each individual molecule absorbs more energy
$1
$100
When same total amount of energy is
distributed amongst a smaller number of
molecules, each individual molecule receives
more energy.
10 energy units
10 Energy Units
split 5 ways
split 2 ways
Link to video
Higher Temp, each
molecule has more KE
Finish the rest of 12-3
CONTAINER #1 = $500
CONTAINER #1
CONTAINER #2 = $5000
CUP
BUCKET OF WATER
#1: 24 ÷ 6 = 4
4
4
24
12
#2: 12 ÷ 3 = 4
TEMPERATURES ARE SAME; SAME AVERAGE KINETIC ENERGY
CONTAINER #1 HAS STORED MORE TOTAL HEAT 24 vs 12
#1: 300 ÷ 3 = 100
#2: 500 ÷ 10 = 50
100
50
300
500
CONTAINER #1 ; HIGHER AVERAGE KINETIC ENERGY
CONTAINER #2 HAS STORED MORE TOTAL HEAT 500 vs 300
Do all objects absorb and store heat in the
same way? If both cars are left in the sun in the
senior lot for the same amount of time will the
interiors be at the same temperature at lunch
time?
Black Porsche
White Porsche
Do both of these materials
store heat in the same way?
Water
Rock
“s” is Specific Heat Capacity
• s = amount of energy
needed to raise the temp of 1 g
of a substance by 1ºC.
• H2O:
Rock:
s = 4.18 J/gºC
s = 0.6 J/gºC
• How much energy in J must be added to 1 g of
H2O to raise its temp by 1 oC? 4.18 J
• How much energy in J must be added to 1 g of
rock to raise its temp by 1 oC? 0.6 J
HW 12-3, #8
OCEAN (4.18 J/ goC)
ROCK (0.6 J/ goC)
• MORE energy must
be absorbed to raise
T.
• MORE energy must
be lost to lower T.
• LESS energy must
be absorbed to raise
T.
• LESS energy must
be lost to lower T
HW 12-3, #9
COPPER (0.13 J/ goC)
ALUMINUM (0.89 J/ goC)
• Stores less energy
when raising Temp;
• Releases less
energy when
lowering Temp
• Stores more energy
when raising Temp;
• Releases more
energy when
lowering Temp
Calorimeter – an insulated
container that does not allow
heat to enter or leave
ALUMINUM HAS HIGHER HEAT CAPACITY; Al HAS MORE
HEAT ENERGY STORED at 100 oC THAN Cu. WATER
ABSORBS MORE HEAT FROM Al AND REACHES HIGHER
FINAL TEMP.
Copy at bottom of p. 128; write
equation on top of 129.
Q = s×m×ΔT
• Q = heat absorbed or lost
+Q if absorbed, -Q if lost
• m = mass in grams
• ΔT = temperature change:
Δ T = Tfinal – Tinitial
• s = specific heat capacity
units of s are J/gºC
EXAMPLE PROBLEM
• Calculate J of energy required to heat 454 g of water from
5.4oC to 98.6 oC. s of H2O = 4.18 J/g oC.
• Q = ? or x
• m = 454 g
• Ti = 5.4 oC ; Tf = 98.6 oC ; ∆T = Tf - Ti = 98.6 -5.4
=93.2 oC
• Q = sm ∆T = (4.18 J/ g oC)(454 g)(93.2 oC)
• Q = 1.77 x 105 J