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Thermodynamics
Brief walk-through of temperature, heat, and energy transfer
Temperature
Temperature: A measure of how hot (or cold) something is Specifically, a measure of the average kinetic
energy of the particles in an object.
Thermometers
Thermometer: an instrument that measures and indicates temperature
Bimetallic strip
a. b.
c.
Temperature Scales
Kelvin International System (SI) of measurement
Fahrenheit Based on 30 being the freezing point of water and
100 as the normal body temperature (later revised to 32 and 98.6)
Celsius Based on water freezing at 0°C, boiling at 100°
Kelvin and Absolute Zero
The Kelvin scale is based on absolute zero Absolute Zero: the temp at which molecular
movement stops 0 K on the Kelvin scale = -273.16ºC
Tc + 273 = TK
What is heat?
Heat: ENERGY created by the motion of atoms and molecules. All matter has heat.
Put energy into a system and it heats up. Take energy away and it cools down.
(applet) The more kinetic energy the particles of a
substance have, the greater the temperature of the object
Temperature and Thermal Energy
The wide range of temperatures present in the universe is shown in the figure.
Temperatures do not appear to have an upper limit. The interior of the Sun is at least 1.5×107°C. Temperatures do, however, have a lower limit.
Temperature Limits
Energy Transfer
The transfer of heat is normally from a high temperature object to a lower temperature object.
1) Conduction
Thermal Conduction: the transfer of heat within a substance, molecule by molecule.
2) Convection
Convection: the movement of matter due to the differences in density that are caused by temp. variations applet
3) Radiation
Radiation: the energy that is transferred as electromagnetic waves, Doesn’t need matter
Most radiation comes from the sun
Conductor vs. Insulator
Conductor: any material through which energy can be transferred as heat
Insulator: poor conductors
Energy conversion
Mechanical energy is converted into thermal energy whenever you bounce a ball. Each time the ball hits the ground, some of the energy of the ball's motion is converted into heating up the ball, causing it to slow down at each bounce
Heat Transfer
The specific heat of a material is the amount of energy that must be added to the material to raise the temperature of a unit mass by one temperature unit.
Heat Transfer = Q = mC(Tf – Ti) Where:
m = mass of object C = specific heat of object; T = temperature in Kelvins
Liquid water has a high specific heat compared to the specific heat of other substances.
A mass of 1 kg of water requires 4180 J of energy to increase its temperature by 1 K. The same mass of copper requires only 385 J to increase its temperature by 1 K.
Specific Heat
Example
A 5.10 kg cast iron skillet is heated on the stove from 295 K to 450 K. How much heat had to be transferred to the iron? The specific heat of iron is 450 J/kg*K M = 5.10 kg; ∆T = 450 – 295 = 155 K
Q = 5.10(450)(155) = 3.6 E 5 Joules
A system is composed on two model blocks at different temps that are initially separated. When they are brought together, heat flows from the hotter block to the colder block. Total energy remains constant.
Conservation of Energy
Conservation of Energy
Qbefore = Qafter
mACaTi + mBCbTi = mACaTf + mBCbTf
Example – Your Turn
A calorimeter contains 0.50 kg of water at 15 degrees Celsius. A 0.040 kg block of zinc at 115 degrees Celsius is placed in the water. What is the final temperature of the system?
Temperature and Thermal Energy
Transferring Heat in a Calorimeter
A calorimeter contains 0.50 kg of water at 15°C. A 0.040-kg block of zinc at 115°C is placed in the water. What is the final temperature of the system?
Section
12.1
Step 1: Analyze and Sketch the Problem
Temperature and Thermal Energy
Transferring Heat in a Calorimeter
Section
12.1
Transferring Heat in a Calorimeter
Let zinc be sample A and water be sample B.
Section
12.1 Temperature and Thermal Energy
Sketch the transfer of heat from the hotter zinc to the cooler water.
Transferring Heat in a Calorimeter
Identify the known and unknown variables.
Temperature and Thermal EnergySection
12.1
Unknown:Tf = ?
Known:
mA = 0.040 kg
CA = 388 J/kg·ºC
TA = 115 ºC
mB = 0.500 kg
CB = 4180 J/kg·ºC
TB= 15.0 ºC
Step 2: Solve for the Unknown
Temperature and Thermal Energy
Transferring Heat in a Calorimeter
Section
12.1
Transferring Heat in a Calorimeter
Determine the final temperature using the following equation.
Temperature and Thermal EnergySection
12.1
m C T +m C TT =
m C +m CA A A B B B
fA A B B
Transferring Heat in a Calorimeter
Substitute mA = 0.040 kg, CA = 388 J/kg·ºC, TA = 115 ºC, mB = 0.500 kg, CB = 4180 J/kg·ºC, TB= 15.0 ºC.
Temperature and Thermal EnergySection
12.1
Tf
0.040 kg 388 J/kg·ºC 115 ºC + 0.500 kg 4180 J/kg·ºC 15.0 ºC=
0.040 kg 388 J/kg·ºC + 0.500 kg 4180 J/kg·ºC
Tf = 16ºC
Step 3: Evaluate the Answer
Temperature and Thermal Energy
Transferring Heat in a Calorimeter
Section
12.1
Temperature and Thermal Energy
Are the units correct?
Temperature is measured in Celsius.
Is the magnitude realistic?
The answer is between the initial temperatures of the two samples, as is expected when using a calorimeter.
Transferring Heat in a Calorimeter
Section
12.1
Transferring Heat in a Calorimeter
The steps covered were:
Temperature and Thermal Energy
Step 1: Analyze and Sketch the Problem
Let zinc be sample A and water be sample B.
Sketch the transfer of heat from the hotter zinc to the cooler water.
Section
12.1
Transferring Heat in a Calorimeter
The steps covered were:
Temperature and Thermal Energy
Step 2: Solve for the Unknown
Determine the final temperature using the following equation.
Step 3: Evaluate the Answer
Section
12.1
Temperature and Thermal Energy
Animals can be divided into two groups based on their body temperatures.
Most are cold-blooded animals whose body temperatures depend on the environment.
The others are warm-blooded animals whose body temperatures are controlled internally.
That is, a warm-blooded animal’s body temperature remains stable regardless of the temperature of the environment.
Calorimetry: Measuring Specific Heat
Section
12.1
Changes of State
At normal atmospheric pressure, water boils at 373 K. The thermal energy needed to vaporize 1 kg of a liquid is called the heat of vaporization.
Changes of State and the Laws of Thermodynamics
The figure below diagrams the changes of state as thermal energy is added to 1.0 g of water starting at 243 K (ice) and continuing until it reaches 473 K (steam).
Changes of State
Section
12.2
Changes of State and the Laws of Thermodynamics
This can be observed between points B and C in the figure, where the added thermal energy melts the ice at a constant 273 K.
Because the kinetic energy of the particles does not increase, the temperature does not increase between points B and C.
Melting Point
Section
12.2
Changes of State and the Laws of Thermodynamics
Once a solid is completely melted, there are no more forces holding the particles in the solid state.
Adding more thermal energy again increases the motion of the particles, and the temperature of the liquid rises.
Boiling Point
Section
12.2
In the diagram, this process occurs between points C and D.
Changes of State and the Laws of Thermodynamics
The amount of energy needed to melt 1 kg of a substance is called the heat of fusion of that substance.
Heat of fusion
Section
12.2
The added energy causes a change in state but not in temperature. The horizontal distance in the figure from point B to point C represents the heat of fusion.
Changes of State and the Laws of Thermodynamics
At normal atmospheric pressure, water boils at 373 K.
The thermal energy needed to vaporize 1 kg of a liquid is called the heat of vaporization. For water, the heat of vaporization is 2.26106 J/kg.
The distance from point D to point E in the figure represents the heat of vaporization. Every material has a characteristic heat of vaporization.
Heat of Vaporization
Section
12.2
Changes of State and the Laws of Thermodynamics
Between points A and B, there is a definite slope to the line as the temperature is raised. This slope represents the specific heat of the ice.
The slope between points C and D represents the specific heat of water, and the slope above point E represents the specific heat of steam.Note that the slope for water is less than those of both ice and steam. This is because water has a greater specific heat than those of ice and steam.
Heat of Vaporization
Section
12.2
Changes of State and the Laws of Thermodynamics
The values of some heats of fusion, Hf, and heats of vaporization, Hv, are shown in the table below.
Heat of Vaporization
Section
12.2
Heat
Suppose that you are camping in the mountains. You need to melt 1.50 kg of snow at 0.0°C and heat it to 70.0°C to make hot cocoa. How much heat will be needed?
Section
12.2 Changes of State and the Laws of Thermodynamics
Step 1: Analyze and Sketch the Problem
Heat
Section
12.2 Changes of State and the Laws of Thermodynamics
Heat
Sketch the relationship between heat and water in its solid and liquid states.
Section
12.2 Changes of State and the Laws of Thermodynamics
Heat
Sketch the transfer of heat as the temperature of the water increases.
Section
12.2 Changes of State and the Laws of Thermodynamics
Heat
Identify the known and unknown variables.
Section
12.2
Unknown:Qmelt ice = ?Qheat liquid = ?Qtotal = ?
Known:
mA = 1.50 kg
Ti = 0.0 ºC
C = 4180 J/kg· ºC
Hf = 3.34×105 J/kg
mB = 0.500 kg
Tf= 70.0 ºC
Changes of State and the Laws of Thermodynamics
Step 2: Solve for the Unknown
Heat
Section
12.2 Changes of State and the Laws of Thermodynamics
Heat
Calculate the heat needed to melt ice.
Section
12.2 Changes of State and the Laws of Thermodynamics
Qmelt ice = mHf
Heat
Substitute mA = 1.50 kg, Hf = 3.34×105 J/kg.
Section
12.2 Changes of State and the Laws of Thermodynamics
Qmelt ice = (1.50 kg) (3.34×105 J/kg) = 5.01×105 J = 5.01×102 kJ
Heat
Calculate the temperature change.
Section
12.2 Changes of State and the Laws of Thermodynamics
ΔT = Ti – Tf
Heat
Section
12.2 Changes of State and the Laws of Thermodynamics
Substitute Tf = 70.0 ºC, Ti = 0.0 ºC
ΔT = 70.0 ºC – 0.0 ºC = 70.0 ºC
Heat
Calculate the heat needed to raise the water temperature.
Section
12.2 Changes of State and the Laws of Thermodynamics
Qheat liquid = mCΔT
Heat
Section
12.2 Changes of State and the Laws of Thermodynamics
Substitute mA = 1.50 kg, C = 4180 J/kg· ºC, ΔT = 70.0 ºC.
Qheat liquid = (1.50 kg) (4180 J/kg· ºC) (70.0 ºC) = 4.39×105 J = 4.39×102 kJ
Heat
Calculate the total amount of heat needed.
Section
12.2 Changes of State and the Laws of Thermodynamics
Qtotal = Qmelt ice + Qheat
liquid
Heat
Section
12.2 Changes of State and the Laws of Thermodynamics
Substitute Qmelt ice = 5.01×102 kJ, Qheat liquid = 4.39×102 kJ.
Qtotal = 5.01×102 kJ + 4.39×102 kJ
= 9.40×102 kJ
Step 3: Evaluate the Answer
Heat
Section
12.2 Changes of State and the Laws of Thermodynamics
Heat Engines
Section
12.2 Changes of State and the Laws of Thermodynamics
Click image to view the movie.
Changes of State and the Laws of Thermodynamics
When the automobile engine is functioning, the exhaust gases and the engine parts become hot.
As the exhaust comes in contact with outside air and transfers heat to it, the temperature of the outside air is raised.
In addition, heat from the engine is transferred to a radiator.
Outside air passes through the radiator and the air temperature is raised.
Heat Engines
Section
12.2
Changes of State and the Laws of Thermodynamics
All of this energy, QL, transferred out of the automobile engine is called waste heat, that is, heat that has not been converted into work.
When the engine is working continuously, the internal energy of the engine does not change, or ΔU = 0 = Q – W.
The net heat going into the engine is Q = QH – QL. Thus, the work done by the engine is W = QH – QL.
Heat Engines
Section
12.2
Changes of State and the Laws of Thermodynamics
If heat engines completely converted thermal energy into mechanical energy with no waste heat, then the first law of thermodynamics would be obeyed.
However, waste heat is always generated, and randomly distributed particles of a gas are not observed to spontaneously arrange themselves in specific ordered patterns.
The Second Law of Thermodynamics
Section
12.2
Carnot’s result is best described in terms of a quantity called entropy, which is a measure of the disorder in a system.
Changes of State and the Laws of Thermodynamics
In the nineteenth century, French engineer Sadi Carnot studied the ability of engines to convert thermal energy into mechanical energy.
He developed a logical proof that even an ideal engine would generate some waste heat.
The Second Law of Thermodynamics
Section
12.2
Changes of State and the Laws of Thermodynamics
The change in entropy, ΔS, is expressed by the following equation, in which entropy has units of J/K and the temperature is measured in kelvins.
The Second Law of Thermodynamics
Section
12.2
The change in entropy of an object is equal to the heat added to the object divided by the temperature of the object.
QS
TΔ =Change in Entropy
Changes of State and the Laws of Thermodynamics
The increase in entropy and the second law of thermodynamics can be thought of as statements of the probability of events happening.
The Second Law of Thermodynamics
Section
12.2
The figure illustrates an increase in entropy as food-coloring molecules, originally separate from the clear water, are thoroughly mixed with the water molecules over time.