Temperature, Heat, and the First Law of Thermodynamics

Temperature, Heat, and the First Law of Thermodynamics

Chapter 18 Temperature, Heat, and the First Law of Thermodynamics Copyright 2014 John Wiley & Sons, Inc. All rights reserved. Outline -Temperature -The Celsius, Fahrenheit and Kelvin -Scales -Thermal Expansion -Absorption of Heat

Heat capacity specific heat, latent heat, thermal equilibrium 2014 John Wiley & Sons, Inc. All rights reserved. Temperature p(58) 18-1

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 Kelvin scale, which is marked in units called kelvins. 2014 John Wiley & Sons, Inc. All rights reserved. Temperature p(59) 18-1 Two bodies are in thermal equilibrium if they are at the same temperature throughout and therefore no heat will flow from one body to the other. The Zeroth Law of Thermodynamics 2014 John Wiley & Sons, Inc. All rights reserved.

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, change in pressure, change in electrical resistance of a wire with temperature change. 2014 John Wiley & Sons, Inc. All rights reserved.

Temperature p(59) 18-1 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 inserted into the well of the cell. 2014 John Wiley & Sons, Inc. All rights reserved. A triple-point cell The Celsius and Fahrenheit Scales p(62) 18-2 The Celsius temperature scale is defined by with T in kelvins.

The Fahrenheit temperature scale is defined by The Kelvin,Celsius,and Fahrenheit temperature scales compared. 2014 John Wiley & Sons, Inc. All rights reserved. Kelvin scale Absolute zero is the lowest possible temperature where nothing could be colder and no heat energy remains in a substance.

2014 John Wiley & Sons, Inc. All rights reserved. :Example Converting to kelvin* solution Converting * solution 2014 John Wiley & Sons, Inc. All rights reserved.

Eexample Convert solution Conversion solution 2014 John Wiley & Sons, Inc. All rights reserved. Converting

Converting Example Example Converting Converting There are two steps to convert

from Kelvin to Fahrenheit: 2014 John Wiley & Sons, Inc. All rights reserved. Thermal Expansion p(64) 18-3 Linear Expansion All objects change size with changes in temperature. For a temperature change TT, a change TL in any linear dimension (L) is given by in which is the coefficient of linear expansion.

The strip bends as shown at temperatures above this reference temperature. Below the reference temperature the strip bends the other way. Many thermostats operate on this principle, making and breaking an electrical contact as the temperature rises and falls. 2014 John Wiley & Sons, Inc. All rights reserved. Thermal expansion is the tendency of matter to change in volume in response to a change in temperature =11 2014 John Wiley & Sons, Inc. All rights reserved.

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 Fore brass:=19x10-6 /c :Solution L= L T L= 19x10-6 x 2(50-22)=0.001064 m L+L=2+0.001064=2.001064 m 2014 John Wiley & Sons, Inc. All rights reserved.

Thermal Expansion p(65) 18-3 Volume Expansion If the temperature of a solid or liquid whose volume is V is increased by an amount TT, the increase in volume is found to be in which is the coefficient of volume expansion and is related to linear expansion in this way, Answer: (a) 2 and 3 (same increase in height), then 1, and then 4 (b) 3, then 2, then 1 and 4 (identical increase in area)

2014 John Wiley & Sons, Inc. All rights reserved. Sample Problem 18.02 p(62) Thermal expansion of a On a hot day in Las Vegas, an oil trucker loaded 37 000 L of 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/ c, and the coefficient of linear expansion for his steel truck p(6 p(62) 2) tank is 11 x10-6 /c Calculations:

V= (37000) (9.50 x 10-4)(-23) = -808 L Thus, the amount delivered was V (del) = V+V = 37000 808 = 36190 L 2014 John Wiley & Sons, Inc. All rights reserved. The heat capacity is defined as the amount of heat energy needed to raise the temperature of a sample by 1 degree Celsius. J/Co The specific heat capacity is defined as the amount of heat energy needed to

raise 1kg of sample by 1 degree Celsius. where c is called the specific heat capacity or specific heat. Absorption of Heat p(68) 18-4 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 changeTTof the object; that is, in which Ti and Tfare the initial and final temperatures of the object. If the object has

mass m, then, where c is the specific heat of the material making up the object. Answer: Material A has the greater specific heat 2014 John Wiley & Sons, Inc. All rights reserved. Absorption of Heat p(69) 18-4 2014 John Wiley & Sons, Inc. All rights reserved.

It takes 487.5 J to heat 25 grams of copper from 25 C to 75 C. What is the specific heat in Joules/gC? Q= mcTT 487.5 = (25)c(75 - 25 ) 487.5 = (25 )c(50 ) Solve for c: c = 487.5 /(25)(50) c = 0.39 J/gC 2014 John Wiley & Sons, Inc. All rights reserved. 2014 John Wiley & Sons, Inc. All rights reserved.

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