Heat and temperature
April
Vocab
for
review
Celsius
conduction
energy transfer
freezing
heat
Kelvin
law of conservation of energy
melting
phase
potential energy
radiation
specific heat
temperature
work
condensation
convection
evaporation
gas
joules
kinetic energy
liquid
molecular level
phase change
pressure
solid
sublimation
thermal energy
Vocab for mastery
law of entropy
thermal equilibrium
thermodynamics
Non-academic vocab
closed system
co-efficient
component
contraction
decay
deterioration
disorder
efficient
expansion
macroscopic
medium / media
relate directly
Engagement
*
http://jersey.uoregon.edu/vlab/Thermodynamics/index.html
*Thermal equilibrium: the state in which two bodies in physical
contact with each other have identical temperatures.
Engagement activity
EXPLORATION
• Virtual Lab – Determination of Specific Heat
• The amount of heat required to raise the
temperature of a solid body depends on its
change in temperature (DeltaT), its mass (m),
and an intrinsic characteristic of the material
forming the body called specific heat (cp). The
heat is calculated from the equation
• q = cp x m x ΔT
• Historically, heat (q) was measured in terms of
calories. The calorie was defined as the amount
of heat required to raise the temperature of 1
gram of water by 1 ⁰C from 14.5 ⁰C to 15.5 ⁰C at 1
atmosphere pressure. With this definition, the
specific heat of water is 1.00 cal/(g · ⁰C ). The use
of the calorie began before it was established
that heat is a form of energy and 1 calorie is
equivalent to 4.18 J. The joule (J) has become the
more favored unit in recent years. Thus, the units
for cp that we will use are J/(g · ⁰C ). The specific
heat of water is then 4.18 J/(g · ⁰C ).
• PURPOSE:
• To apply the experimental methods of
calorimetry* in the determination of the
specific heat of a metal.
•
*calorimetry: an experimental procedure used to measure the energy transferred
from one substance to another as heat.
• PROCEDURE:
• Perform three trials for EACH material. Record the mass of
the sample, the initial temperature (T1), the final
temperature (T2), and the energy absorbed (in joules).
• Vary the mass of the sample and the length of time heating
for each trial with a given material. In other words, no two
trials for a given material should use both the same mass
and the same flame exposure times.
• At the end of each trial, hit the "Reset" button to prepare
for the next trial.
• http://www.sciencegeek.net/VirtualLabs/SpecificHeatLab.h
tml
RESULT
• Observation and data:
Material____
_______
Trial #1
Trial #2
Trial #3
Mass of
object, m (g)
g
g
g
Initial
temperature
of object (T1)
⁰C
⁰C
⁰C
Initial
temperature
of object (T2)
⁰C
⁰C
⁰C
J
J
J
Energy
absorbed
(Joules)
result
• CALCULATIONS: Show your work! You will
repeat the calculations three times for each
material you test.
• Calculate the temperature change, ΔT, (T2 T1).
• Calculate the specific heat, Cp. Cp = q ÷ m ÷ ΔT
Material____
__________
ΔT, (T1 – T2)
Cp
Trial #1
Trial #2
Trial #3
⁰C
⁰C
⁰C
j/(g · ⁰C)
j/(g · ⁰C)
j/(g · ⁰C)
Temperature scales
• Celsius and Fahrenheit scales
• Tf=9/5 Tc + 32
• Celsius and Kelvin (or absolute) scales
• T= Tc + 273.15
practice
• One day it was -40 ⁰C at the top of Mont Blanc
and -40 ⁰F at the top of Mount Whitney.
Which place was colder?
• Neither (-40 ⁰C =-40 ⁰F)
temperature and thermal equilibrium
• When it is 75⁰F, you feel warm. When it is 45 ⁰F,
you feel cool.
• Air has more energy at 75 ⁰F than it has at 45 ⁰F.
This energy is the kinetic energy of air molecules.
• When two liquids or gases at different
temperatures are mixed, the stable temperature
is between the two temperatures. This is called
thermal equilibrium.
conceptual challenge
• If two cups of hot chocolate,, one at 50 ⁰C and
the other at 60 ⁰ C, are poured together in a
large container, will the final temperature of
the double batch be
• a. less than 50 ⁰ C?
• b. between 50 ⁰F and 60 ⁰ C?
• c. greater than 60 ⁰ C ?
heat and energy
• Why do you feel colder
with the tile floor than
with the wooden floor?
• The tile transfers or
conducts energy faster
than the wood does.
More energy is removed
from your feet in a given
time.
• That’s why you feel colder
with the tile floor.
• Heat always moves
• Heat has the units of
from an object at higher
energy.
temperature to an
• Joule (J)
object at lower
• calorie (cal)=4.186J
temperature.
• Heat is defined as
energy in transit.
Specific heat capacity
• We experience that an object may be hotter
than the other in the same condition.
• That property of objects is known as specific
heat capacity.
• The specific heat capacity is defined as the
energy required to change the temperature of
1 g of a substance by 1 ⁰C.
specific heat capacity
• cp = q / mΔT
• Specific heat capacity = heat energy/(mass x
change in temperature)
• Water: 4.186 (J/g ⁰C)
• Iron: 4.48 (J/g ⁰C)
calorimetry
• You are preparing to take a bath. The coldwater faucet supplies water at 20.0 ⁰C, and
the water from the hot-water faucet is 60.0
⁰C. Each faucet has poured 25.0 kg of water
into the tub. What is the temperature of the
bath?
• 40.0 ⁰C
Calorimetry practice
• You prefer your bath at 30.0 ⁰C. The hot-water
faucet has already poured 20.0 kg of water at
60.0 ⁰C into the tub. How much cold water
(20.0 ⁰C) should you add?
• 60.0kg
Practice
• What is the final temperature when a 3.0 kg gold bar at
99 ⁰C is dropped into 0.22 kg of water at 25 ⁰C?
• 1. set the final temperature as T.
• 2. the gold bar loses heat and water gains heat.
• 3. loss of heat is q =cp mΔT = 0.129x3000x(99-T)
• 4. gain of heat is q= cp mΔT =4.186x220x(T-25)
• 5. Loss of heat is equal to gain of heat.
• 0.129x3000x(99-T) = 4.186x220x(T-25)
• 387(99-T)=920.92(T-25)
• T=47 ⁰C
Homework
•
•
•
•
Page 374
Practice 10C
Questions 2, 3 and 7
For the specific heat
capacities of substances,
refer to the table 10-4 in
page 372. Be aware that
the specific heat in the
textbook is measured in
J/kg ⁰C.
•
•
•
•
Answers
18.0 ⁰C
79 ⁰C
135 g
Phase change
1. What is the state of substances during Leg “A”?
2. What is the state of substances during Leg “B”?
3. What is the state of substances during Leg “D”?
•
•
•
•
1. Leg “A”: temperature of
ice changes as energy is
added.
2. Leg “B”: temperature of
ice does not change. Instead, the nature of the
ice changes. The ice-and-water mixture remains
at this temperature until all of the ice melts.
3. Leg “C”: the water’s temperature increases as
energy is added.
•
•
•
•
4. Leg “D”: temperature
stops rising and the water
turns into steam.
5. Leg “E”: the steam’s
temperature increases.
• During Leg “B” and Leg “D,” the energy added or
removed changes the internal energy of the
substance without changing its temp. These
changes in matter are called phase changes.
Phase change
vaporization
condensation
melting
solidification
Phase change
heat of vaporization
2.26x103 J/g
heat of fusion 3.33 x102 J/g
heat of phase change practice
• How much energy is required to change a 42 g ice cube
from ice at -11 ⁰C to steam at 111 ⁰C?
• (heat of vaporization is 2.26x103 J/g, and heat of fusion is
3.33 x102 J/g)
• 1. at the phase of ice: q=cp mΔT=4.186x42x11
• 2. at the state of a mixture of ice and water: heat of fusion
x mass= 3.33 x102 J/g x 42g
• 3. at the phase of water: q=cp mΔT =4.186x42x100
• 4. at the state of a mixture of water and steam: heat of
vaporization x mass = 2.26x103 J/g x 42g
• 5. at the phase of steam: q=cp mΔT =4.186x42x11
• Answer: 1.30 x 105 J
Draw a phase change diagram
• Ice at -20⁰C is heated to steam at 133.0 ⁰C.
Homework practice
• Page 381
• #3, #4 and #5
•
•
•
•
Answers
1.415 x104J
1.354 x107J
76.2 ⁰C