Tue Nov 17
• Assign 13 - Friday
Today:
• Heat/Calorimetry
• Latent Heats and Phase Changes
Q = cmΔT
If NO PHASE CHANGE!
c – specific heat capacity (J/kg ºC)
Material and phase specific
ΔT - ºC or K
Water
4186
Ice
2090
Water Vapor
2010
Aluminum
900
Copper
387
How much heat is required to increase the temperature of 5.0 kg of
water by 10 °C?
(1) 2.09 x 103 J
(4) 2.09 x 105 J
(2) 8.37 x 103 J
(5) 8.37 x 105 J
Q = mcΔT
Q = (5.0k) (4186 J/°C kg) (10 °C)
= 2.09 x105 J
(3) 9.28 x 103 J
(6) 9.28 x 105 J
Calorimetry
QLOST = QGAINED
ΔT always positive in this case
(not always real Δ)
(will be Tinitial – Tfinal for QLOST)
• “Accounting”
• Careful with algebra
• Reality check
NOTE: System (stuff we care about) isolated. No energy in or out
from outside system. Otherwise we would need to factor in external
energy source or drain.
Start with two beakers of 200g of water at 20°C.
There are two blocks, each with a mass of 200g and a temperature of
80°C.
One block is aluminum and one is copper. Place one block in each
beaker and let them reach thermal equilibrium.
Which beaker has the higher temperature in the end?
(1) the one with Aluminum
(2) the one with Copper
(3) both the same final temp
Al: TF = 30.6º
Cu: TF = 25.1º
Higher c for aluminum. Will increase temp of water more for
each degree of temperature lost by cylinder.
QLOST = QGAINED
mAl cAl ΔTAl = mW cW ΔTW
(0.2kg)(900 J/kgºC)(80º-TF ) = (0.2kg)(4186 J/kgºC)(TF -20º)
TF = 30.6º
A 1.50-kg block of aluminum at an initial temperature of 20.0°C is
placed over a 150-W electric heater (a 150-W electric heater puts out
150 J of energy every second) for 300 seconds. What is the increase
in temperature of the aluminum block presuming all the energy goes
into the block in the form of heat?
(1) 10.0°C
(5) 41.5°C
(2) 13.3°C
(6) 53.3°C
(3) 20.0°C
(7) 58.2°C
(4) 33.3°C
(8) 64.1°C
Q = mcΔT
ΔT = Q/mc
Q = Power * time = 150W*300s = 45000J
ΔT = 45000J/(1.5kg*900 J/kg°C) = 33.3
Want ΔT, not TFINAL so don’t have to do anything else.
Phases
•
•
•
•
Solid – retains shape
Liquid – conforms to shape of container, preserves volume
Gas – expands to fill volume
Plasma – electrons and ions – gas-like
PhET
Latent Heat – Phase Change
• Changing phase – requires or produces energy
• Related to bonds
– Ice → liquid water (melt or fuse)
Requires Energy
– Liquid → ice (freeze)
Releases Energy
• No ΔT during change
• Two phases can exist at same T
Q = mL
• L is latent (hidden) heat
• Same value L for inverse transitions
Water:
LFUSION = 33.5x104 J/kg
LVAPOR = 22.6x105 J/kg
cICE = 2090 J/kg ºC
cSTEAM = 2010 J/kg ºC
cWATER = 4186 J/kg ºC
Water:
LFUSION = 33.5x104 J/kg
LVAPOR = 22.6x105 J/kg
cICE = 2090 J/kg ºC
cSTEAM = 2010 J/kg ºC
cWATER = 4186 J/kg ºC
How much energy is required to melt 0.5kg of ice?
(1) 1045 J
(2) 2090 J
(3) 2093 J
(4) 4186 J
(5) 16.7x104 J (6) 33.5x104 J
(7) 11.3x105 J (8) 22.6x105 J
Phase transition – use latent heat of fusion
Q = m LFUSION = (0.5kg) (33.5x104 J/kg)
Latent Heat – Phase Change
Liquid
Gas
Liquid
Energy
Q = cmΔT
Gas
Energy
Q = mL
Build the story, then translate into the math
Water:
LFUSION = 33.5x104 J/kg
LVAPOR = 22.6x105 J/kg
cICE = 2090 J/kg ºC
cSTEAM = 2010 J/kg ºC
cWATER = 4186 J/kg ºC
Temp ºC
Consider 1kg of material A (Blue) and material B (Green). Which
requires the most energy to change from the liquid to the gas
phase?
B
A
(1) A
gas
(2) B
(3) Both the same
liquid
Solid
Heat
An 500-g aluminum cylinder at T=20 °C is heated using steam at
T=140°C. What is the minimum mass of the steam required to raise
the temperature of the aluminum cylinder by 40°C?
QGAIN = QLOST
(heat Alum) = (cooling steam) + (condensing steam) + (cooling water)
mAl cAl ΔTAl = (mScSΔTS) + (mS LV) + (mS cW ΔTW )
(0.5)(900)(60-20) = mS2010(140-100) + mS2.2x106 + mS4186(100-60))
mS = 7.18 x10-3 kg
Caution points:
• mass of water is same as mass of steam
• BUT specific heats are not the same
Example:
A slug of copper (0.2kg) at 800 ºC is immersed in 0.4kg of water at
80 ºC.
The final temp is 100 ºC. Part of the water boiled off as steam.
What was the mass of the water converted to steam?
Find Q lost by copper
Find Q to heat water to 100 ºC
Rest of Q goes into evaporation
QLOST = QGAINED
Cool Copper = Warm up water + convert some to steam
mCu cCu ΔTCu = mW cW ΔTW + mSTEAM LVaporization
(0.2)(387)(800-100) = (0.4)(4186)(100-80) + mSTEAM (2.26x106)
mSTEAM = 9.16 x10-3 kg
Suppose the latent heat of vaporization of water were one-tenth its
actual value. All other things being equal, the time it would take for a pot of water
to boil completely away would be:
(1) a longer time.
(2) the same time.
(3) a shorter time.
Less energy required. Presuming that stove provides constant
power, takes shorter time to provide all off the necessary energy.
Bullet A at a temp of 37ºC is placed on a block of ice at 0 ºC. Bullet B at a temp of 0ºC and a speed of 200 m/s is stopped by a block
of ice at 0 ºC. Both bullets are the same mass. Which bullet melts
more ice?
cbullet = 128 J/kg ºC
1. Bullet A
2. Bullet B
3. Not enough information
QLOST,A = m cBULLET ΔT
= m (128)(37)
= 4736 m
WB= ΔKE = (1/2)m v2
= 0.5 m (200)2
= 20000 m
A 2.0 kW heater runs for 3 minutes. How much
energy does it produce in Joules?
(1) 6 J
(4) 667 J
(2) 11.1 J
(5) 6000 J
(3) 360J
(6) 360000 J
Energy = Power * Time
Energy = 2000W * 180 seconds = 360000J
Example:
A 400-W "immersion heater" is used to heat 1.25x10-4 m3 of water.
How long does it take to increase the temperature by 30º C?
Power = Energy/Time
Find how much energy is required (heat)
Then find time
Need density to find mass of water
m = ρV = 1000 kg/m3 · 1.25x10-4 m-3 = 0.125 kg
P = ΔE/Δt
Δt=ΔE/P
Need energy ΔE = Q = mcΔT = (0.125kg) (4186J/kg ºC)(30º)
ΔE = 15700 J
Δt=ΔE/P = 15700J/400W = 39s
You have an ice chest filled with 7.0 kg of ice at -1°C sitting in the
trunk of your car. Over the period of two days it melts and warms up
to 27°C. On average, what is the energy per second leaking into the
ice chest?
Q = (warm ice) + (melt ice) + (warm water)
Q= (mciΔTi)+ (mi LF) + (mw cW ΔTW )
Q = (7kg)2090(1C°) + (7kg)33.4x104 + (7kg)4186(27-0)
Q = 1.46x104 J + 2.338x106 J + 7.91x105 J
Q = 3.14 x106 J
If only 5% of the energy produced by the body goes
into an action, and 200J of energy is produced by the
body, how much energy goes into the action?
(1) 5 J
(4) 100 J
(2) 10 J
(5) 190 J
(3) 40J
(6) 195 J
200J * 0.05 = 10 J
So that would be the amount of energy that
could go into lifting a box.
Ideal Gases and Kinetic Theory
• Gas
– fluid
– expands to fill container
• Ideal Gas
– Low enough density that any interaction basically elastic
collision between two molecules
• State: (state variables)
– P, V, T, N
– U (internal energy), entropy, …
• Goal:
– Relationship between variables
– Develop a microscopic model
PhET http://phet.colorado.edu
http://mc2.cchem.berkeley.edu/Java/molecules/
Atoms and Molecules
If you have 48 doughnuts, how many dozen doughnuts do you have?
• A mole is like a dozen, but more
N
• Tiny, lots of them
n=
NA
• Counting:
– 1 mole = NA molecules
Mass per mole
mparticle =
– NA = 6.02 x 1023 (Avogadro constant)
• Mass:
– proton: 1.67 x 10-27 kg
– Periodic table: grams per mole
– same as atomic mass units (amu) or unified mass (u's)
– 1u = 1.6605 x 10-27 kg
– One mole of a 15u molecule has a mass of 15g
NA
What is the mass in grams of 3 moles of CO2?
(1) 3g
(4) 28g
(7) 132 g
(2) 9.3 g
(5) 44 g
(3) 14.7 g
(6) 84 g
Mass = 1 Carbon + 2 Oxygen
= 12u + 2 (16u) = 44u
1 mole of CO2 would be 44 g
3 moles would be 3*44u = 132
Mass in u of atoms:
Hydrogen (H) 1u
Helium (He) 4u
Lithium (Li) 6u
Beryllium (Be) 8u
Boron (Bo)
10u
Carbon (C)
12u
Nitrogen (N) 14u
Oxygen (O)
16u
Relationships: Gas Laws
• Charles' Law
– Const P and n
• Boyle's Law
– Const T and n
Vi V f
=
Ti T f
PiVi = Pf V f
• Ideal Gas Law
– R=8.31 J/mol K (ideal gas constant)
PV = nRT
– kB= R/NA =1.38 x 10-23 J/K (Boltzmann constant)
PV = Nk BT
Use absolute SI units
• If ΔT, then can use Kelvin or Celsuis
• If T is multiplied or divided, must be in K!
• P is PABS. P in Pa (1atm = 1.01x105 Pa)
PABS = PGAUGE + PATM
PV = nRT
Consider two cylinders of the same volume and the same temperature. Container X contains O2.
Container Y contains CO2 (which is heavier per molecule). If there are the same number of moles of gas in each, then (1) Px > Py
(2) Px = Py
(3) Px < Py
Haven't said anything about type of molecule yet.
nRT
P=
V
At STP (1atm, T=0 °C) 1 mole of ANYTHING occupies volume
of 22.4 liters
Consider two cylinders of the same volume and the same temperature. Container X contains O2. Container Y contains CO2 (which is heavier,
per molecule). If the mass of the gas in each container is the same, (1) Px > Py
(2) Px = Py
(3) Px < Py
Fewer molecules of CO2 (lower n). V,T constant
nRT
P=
V
Ideal Gas Law
• V=A*h
• P = F/A
PV = nRT
PV = Nk BT
• Isolate values that are constant
• Suppose constant temperature and pressure, but double
number. What is the ratio of volumes? VAFTER/VBEFORE
PVBEFORE = nBEFORE RT
PVAFTER = n AFTER RT
VBEFORE RT VAFTER
=
=
nBEFORE
P
n AFTER
VAFTER
n AFTER
2n
=
=
=2
VBEFORE nBEFORE
n
Suppose I keep the temperature of a gas constant but triple the
Volume and cut the number of molecules in half. What would
be the ratio of pressures, PAFTER/PBEFORE?
(1) 1/6
(5) 3/2
PBEFORE
(2) 2/3
(6) 2
nRT
=
V
(3) 1/2
(7) 3
(4) 1/3
(8) 6
PAFTER
(
1 2 )nRT
=
3V
1 nRT
=
6 V
Suppose I triple the pressure of a gas and double the volume
while keeping the number of molecules same. What would be
the ratio of temperatures, TAFTER/TBEFORE?
(1) 1/6
(5) 3/2
(2) 2/3
(6) 2
PV
TBEFORE =
nR
(3) 1/2
(7) 3
(4) 1/3
(8) 6
(3P)(2V)
PV
TAFTER =
=6
nR
nR
Kinetic Theory – A Microscopic Model
•
•
•
•
Assume like billiard balls
Pressure from collisions with walls
vRMS most probable speed
All molecules in a gas same temperature
vRMS = v 2
Kinetic Theory – The math
• Took basic Newtonian physics
• Found
2⎛1
• Compare to
PV = N ⎜ mv
3⎝2
KE
2
rms
⎞
⎟
⎠
PV = Nk BT
3
KE = k BT
2
kinetic theory
http://mc2.cchem.berkeley.edu/Java/molecules/
N2 is lighter than O2. Both are lighter than CO2. In this room right now,
which has the highest rms-velocity? (1) N2
(2) O2
(3) CO2(4) All same
All molecules have same average KE.
KE = (1/2) mv2
Smallest mass, largest v
If I seal off the room and raise the temperature by 5 degrees,
what happens to the average KE of the CO2? (1) decreases (2) stays the same
(3) increases
Average KE directly related to T no matter what.
Which gas would be most likely to escape from our
atmosphere?
(1) Hydrogen
(4) Oxygen
(2) Helium
(3) Nitrogen
(5) Carbon Monoxide (6) Carbon Dioxide
The lightest gas per molecule will have the highest
average speeds.