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(1)

Temperature, Heat, and the

First Law of Thermodynamics

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Outline

*

Temperature

*

The Celsius , Fahrenheit and Kelvin

Scales

*

Thermal Expansion

*

Absorption of Heat

Heat capacity

specific heat,

latent heat,

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18-1

Temperature p(58)

Thermodynamics is the study and

application of the thermal energy (often

called the internal energy) of systems. One of the central concepts of thermodynamics is temperature.

Temperature is an SI base quantity related to our sense of hot and cold.

 It is measured with a thermometer, which contains a working substance

with a measurable property, such as length or pressure, that changes in a regular way as the substance becomes hotter or colder.

 Physicists measure temperature on the

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18-1

Temperature p(59)

The Zeroth Law of Thermodynamics

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Thermometer

A thermometer is a device that measures the

temperature of things.

Thermometer works through a change in the physical properties of the temperature change, such as property of extended objects with increasing temperature,

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18-1

Temperature p(59)

Triple Point of Water

• The Triple point of water is the point in which solid ice, liquid

water, and water vapor coexist in thermal equilibrium. (This does not occur at normal atmospheric pressure.)

• By international agreement, the temperature of this mixture has been defined to be 273.16 K. The bulb of a constant-volume gas thermometer is shown

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18-2

The Celsius and Fahrenheit Scales p(62)

• The Celsius temperature scale is defined by

with T in kelvins.

• The Fahrenheit temperature scale is defined by

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Absolute zero is the lowest possible temperature where nothing could be colder and no heat energy remains in a substance.

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Example

:

*

Converting to kelvin

 

solution

*

Converting

solution

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Eexample

• Convert

 

solution

Conversion

solution

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Converting

Example

Converting

There are two steps to convert from Kelvin to Fahrenheit:

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18-3

Thermal Expansion p(64)

• All objects change size with changes in temperature. For a

temperature change

ΔT

, a change

ΔL

in any linear dimension

L

is given by

The strip bends as shown at tempera trues above this reference temperature. Below the reference temperature the strip bends the other way. Many thermo- stats operate on this principle, making and breaking an electrical contact as the

Linear Expansion

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Thermal expansion

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EXAMPLE

Consider a 2 m long brass rod at temperature is 22 °C,

if temperature increased to 50 c˚. what would the length

of rod be ??take the linear expansion of brass is.

Solution

∆L= α L ∆T

∆L= 19x10

-6

x 2(50-22)

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18-3

Thermal Expansion p(65)

• If the temperature of a solid or liquid whose volume is

V

is

increased by an amount

ΔT

, the increase in volume is found to

be

in which β is the

coefficient of volume expansion

and is related to linear expansion in this way,

Volume Expansion

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Thermal expansion of a On a hot day in Las Vegas, an oil trucker loaded 37 000 Lof diesel fuel. He encountered cold weather on the way to

Payson, Utah, where the temperature was 23.0 K lower than in Las Vegas, and where he delivered his entire load. How many liters did

deliver? The coefficient of volume expansion for diesel fuel is 9.50 x 10-4/Co, and the coefficient of linear expansion for his steel truck tank is 11 x 10-6Co

Calculations:

∆V= (37000) (9.50 x 10)(-23) = -808 L

Thus, the amount delivered was

V del = ∆V+V = 37000 – 808 = 36190 L

Sample Problem 18.02

p(62)

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The

specific heat capacity

is defined

as the amount of heat energy needed to

raise 1kg of sample by 1 degree Celsius.

The

heat capacity

is defined as the

amount of heat energy needed to raise

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18-4

Absorption of Heat p(68)

Absorption of Heat by Solids and Liquids

• The heat capacity C of an object is the proportionality constant

between the heat Q that the object absorbs or loses and the resulting temperature changeΔTof the object; that is,

in which Ti and Tfare the initial and final temperatures of the object.

If the object has

mass m, then,

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(20)

It takes 487.5 J to heat 25 grams of copper from 25 °C to

75 °C. What is the specific heat in Joules/g·°C?

Q= mcΔT

487.5 J = (25 g)c(75 °C - 25 °C)

487.5 J = (25 g)

c

(50 °C)

Solve for c:

(21)

References

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