Thermodynamics Brief walk-through of temperature, heat, and energy transfer

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

http://www.colorado.edu/physics/2000/bec/temperature.html

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

http://www.kangwon.ac.kr/~sericc/sci_lab/physics/conduction/convection.html

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?


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



Section

12.1


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.


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



Section

12.1


Determine the final temperature using the following equation.


12.1

m C T +m C TT =

m C +m CA A A B B B

fA A B B


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.


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



Section

12.1


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.


Section

12.1


The steps covered were:



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


The steps covered were:



Determine the final temperature using the following equation.


Section

12.1


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


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


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.


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.


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


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.


Section

12.2


The values of some heats of fusion, Hf, and heats of vaporization, Hv, are shown in the table below.


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


Heat

Section


Heat

Sketch the relationship between heat and water in its solid and liquid states.

Section


Heat

Sketch the transfer of heat as the temperature of the water increases.

Section


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



Heat

Section


Heat

Calculate the heat needed to melt ice.

Section


Qmelt ice = mHf

Heat

Substitute mA = 1.50 kg, Hf = 3.34×105 J/kg.

Section


Qmelt ice = (1.50 kg) (3.34×105 J/kg) = 5.01×105 J = 5.01×102 kJ

Heat

Calculate the temperature change.

Section


ΔT = Ti – Tf

Heat

Section


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


Qheat liquid = mCΔT

Heat

Section


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


Qtotal = Qmelt ice + Qheat

liquid

Heat

Section


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


Heat

Section


Heat Engines

Section


Click image to view the movie.


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


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


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.


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.


Section

12.2


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.


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


The increase in entropy and the second law of thermodynamics can be thought of as statements of the probability of events happening.


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.

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Thermodynamics Brief walk-through of temperature, heat, and energy transfer