Transcript Physics 2053C – Fall 2001

Physics 2053C – Fall 2001
Chapter 14
Heat
Nov. 09, 2001
Dr. Larry Dennis,
FSU Department of Physics
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Specific Heat
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Q = m c T (c is the specific heat)
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Depends on the material.
c = Q/(m T)
Assumes there is no phase change as the result
of the heat.
Q = mL (L is the latent heat of fusion or
vaporization.)
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Heat required to complete a phase change (no
change in temperature included).
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Heating water (35 gm at 30 °C)
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Start at 30 °C, heat until it
has all boiled away.
Step 1: heat to 100 °C (net
change of 70°C.
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Q = m c T
Step 2: heat change required
to convert water to steam.
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Q = m Lv
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Heating water (35 gm at 30 °C)
How much heat is required to
increase the temperature to 100
°C?
Q = m c T
Q = 35 gm * 1 cal/(gm °C ) * (70 °C )
Q = 2450 cal
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Heating water (35 gm at 30 °C)
How much heat is required to boil the
water once it reaches 100 °C?
Q = m Lv
Q = 35 gm * 539 cal/gm
Q = 18865 cal
Note it is much larger than the heat
required to raise the temperature.
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Heating water (35 gm at 30 °C)
If the water is heated at a constant
rate of 100 Watts how long does
it take to boil the water away?
100 W = 100 J/s
4.186 J = 1 cal
So: 100 W = 100 J/s / (4.186 J/cal)
100 W = 23.9 cal/s.
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Heating water (35 gm at 30 °C)
Step 1 (heating from 30 °C to 70 °C):
Recall that
Q = 2450 cal
rate = 23.9 cal/s
Time required = Q/rate
t = 2450 cal/(23.9 cal/s)
t = 102 s
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Heating water (35 gm at 30 °C)
Step 2 (boiling at 100 °C):
Recall that
T
Q = 18865 cal
rate = 23.9 cal/s
Time required = Q/rate
t = 18865 cal/(23.9 cal/s)
t = 789 s
t
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Achieving Equilibrium
Energy
300 g of aluminum at 80 °C is placed placed in a 150 g
aluminum cup containing 400 mL of water at 27 °C.
What is the final temperature of the system?
EAL + EW + Ecup = 0
(mcT)AL + (mcT)W + (mcT)cup = 0
mALcAL(Tf – Ti) + mWCW(Tf-To) + mcupccup(Tf-To) = 0
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Achieving Equilibrium
Energy
300 g of aluminum at 80 °C is placed placed in a 150 g
aluminum cup containing 400 mL of water at 27 °C.
What is the final temperature of the system?
mALcAL(Tf – Ti) + mWCW(Tf-To) + mcupccup(Tf-To) = 0
(mALcAL+ mWCW + mcupccup)Tf = mALcALTi + (mWCW + mcupccup)To
Tf = (mALcALTi + (mWCW + mcupccup)To)/ (mALcAL+ mWCW + mcupccup)
Tf = (.3*900*80 + (.4*4186 + .15*900)*27)/ (.3*900 + .4*4186 + .15*900)
Tf = 33.88 °C
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Achieving Equilibrium
Energy
300 g of aluminum at 80 °C is placed placed in a 150 g aluminum cup
containing 400 mL of water at 27 °C. What is the final temperature of the
system?
Energy lost by 80 °C aluminum block = mcT for the aluminum
= .3 kg * 900 J/kg °C * (33.88 - 80)°C
= - 12452 J
Energy gained by aluminum cup = mcT for the cup
= .15 kg * 900 J/kg °C * (33.88 - 27)°C
= 929 J
Energy gained by water in cup = mcT for the water.
= .4 kg * 4186 J/kg °C * (33.88 - 27)°C
= 11523 J
Total change in energy = -12452 + 11523 + 929 = 0
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CAPA 6
A 169 g piece of lead is heated to 80 °C and then dropped into a
calorimeter containing 515 g of water that initially is at 22 °C.
Neglecting the heat capacity of the container, find the final
equilibrium temperature of (in °C) of the lead and water.
EL + EW = 0  (mcT)L + (mcT)W = 0
mLcL(Tf –TiL) + mWcW(Tf-TiW) = 0
(mLcL + mWcW)Tf = mWcWTiW + mLcLTiL
Tf = (mWcWTiW + mLcLTiL )/(mLcL + mWcW)
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CAPA 6
A 169 g piece of lead is heated to 80 °C and then dropped into a
calorimeter containing 515 g of water that initially is at 22 °C.
Neglecting the heat capacity of the container, find the final
equilibrium temperature of (in °C) of the lead and water.
Tf = (mWcWTiW + mLcLTiL )/(mLcL + mWcW)
Tf = (.169*130*80 + .515*4186*22 )/(.169*130+.515*4186)
Tf = 22.6 °C
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CAPA 8
A lead ball, with an initial temperature of 25 °C is released from a
height of 120 m. It does not bounce when it hits a hard surface.
Assume all the energy of the fall goes into heating the lead. Find
the temperature of the ball after it hits.
Change in Kinetic Energy + Change in Internal Energy = 0
-Mgh + McT  -Mgh + Mc(Tf – Ti) = 0
Mgh/Mc = (Tf – Ti)  (Tf – Ti) = 9.8*120/(130) = 9.05 °C
Tf = Ti +9. 05 °C = 34.05 °C
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CAPA 10 & 11: Heat Transfer by conduction.
A mountain climber wears down clothing 3 cm thick with a total
surface area of 1.75 m2. The temperature at the surface of the
clothing is –19 °C and at the skin is 34.5 °C. Determine the rate of
heat flow by conduction through the clothing assuming it is dry and
that the thermal conductivity, k, is that of down.
Q/t = kA(T1-T2)/L
Q/t = 0.025 W/(m°C)*1.75 m2*(34.5 - -19)°C /0.03m
Q/t = 78 W
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CAPA 10 & 11: Heat Transfer by conduction.
Determine the rate of heat flow by conduction through the clothing
assuming the clothing is wet, so that k is that of water and the jacket
is matted down to 0.5 cm thickness.
Q/t = kA(T1-T2)/L
Q/t = 0.56 W/(m°C)*1.75 m2*(34.5 - -19)°C /0.005m
Q/t =10486 W
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Quiz #9 – Heat & Energy
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Specific heats, latent heats, calorimetry.
Rate of heating.
Energy changes and energy balance in
heat transfer.
Note the topics that are not included:
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Convection, conduction and radiation.
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Quiz #9 – Heat & Energy
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Sample Questions.
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Chap. 14: 2, 3, 5
Sample Problems.
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Chap. 14: 19, 21,25 & CAPA 2 - 7
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Next Time
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Quiz on Chapter 14
Begin Chapter 15
Please see me with any questions or
comments.
See you on Wednesday.
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