Chapter 12

Temperature and Heat

Temperature:

The temperature of an object is a measure of how cold or hot that object is.   More precisely, the temperature of an object is a measure of the average kinetic energy of the atoms and molecules of that object.  A hotter object has faster molecular vibrations.  Temperature is a scalar quantity.

Temperature Scales:

Four temperature scales are commonly used:  Fahrenheit and Rankin as well as Celsius and Kelvin.

Fahrenheit and Celsius are regular scales.  Rankin and Kelvin are the "absolute scales."

Fahrenheit and Celsius Scales:

If an unmarked thermometer is placed in a mixture of ice and water, the liquid level in the thermometer goes down and comes to stop at a certain level.  As long as there is ice to melt, the level remains the same.  When all of the ice is molten, then the liquid level starts going up.  If the same unmarked thermometer is placed in pure water and heated, the mercury level in it keeps going up until water comes to boil.  During the boiling process, the level remains constant again until there is no more water to evaporate.  Celsius named these two fixed points as "0" and "100" and made a thermometer.  The "0" and "100" are decided for experiments performed at ocean level where the air pressure is one atmosphere.

Experiments have shown that the temperature of a pure substance remains constant during freezing and melting processes (when phase changes).   The heat absorbed or given off during melting or freezing contributes to phase change only,  either from solid to liquid or from liquid to solid.  The same is true for evaporation and condensation processes during which temperature remains constant.

The important point here is that the melting (also, freezing) point of water is used to mark the "0" on the Celsius scale and the evaporation (also, condensation) point of water is used to mark the "100" on the Celsius scale, of course, both performed at one atmosphere of air pressure.

"0" on Celsius scale corresponds to "32" on Fahrenheit scale and "100" on Celsius scale corresponds to "212" on Fahrenheit scale.  This means that pure water freezes at 32oF and boils at 212o F if the air pressure is one atmosphere.

Comparison Between Celsius and Fahrenheit Scales:

Although most calculators do the conversions between these two scales immediately, it is worth knowing how one measure can be converted to the other

The "0" on Celsius is equivalent to "32" on Fahrenheit, and the "100" on Celsius is equivalent to "212" on Fahrenheit.  Corresponding to every "C " on Celsius, there is an "F" on Fahrenheit as shown. To develop a formula that calculates "C" for a given "F" and vice versa, we may equate the segment ratios on both scales as shown below:

Fig. 1

Example 1:  The Fahrenheit scale reads 77oF, what is the reading on the Celsius scale?

Solution: Using the formula: oC = (5/9)[F - 32] ;  oC = (5/9)[77-32] = 25

Example 2:  A temperature difference of ΔC = 24oC is measured between two points on Celsius scale.  How much is this difference in Fahrenheit scale?

Solution:  100oC difference on the Celsius scale corresponds to 180oF difference on the Fahrenheit scale.  Using a proportion, the difference in Fahrenheit is

ΔF/ΔC  = 180o/100o ;     ΔF/24o = 9/5;     ΔF = 24o (1.8);    ΔF = 43oF.

Example 3:  At what temperature both Fahrenheit and Celsius scales read the same temperature?

Solution:  To be solved by students.

Absolute Scales:  The basis for absolute scales (Kelvin and Rankin scales) is the temperature at which molecular motion comes to stop.  This temperature cannot actually be reached; however, with great cooling, temperatures very close to it have been reached.  Experiments have shown that when a gas is cooled down, its volume decreases.  At constant pressure, the volume decrease for a gas, is proportional to the temperature decrease.  In other words, The ratio ΔV/ΔT remains constant.  That means that if V( the gas volume) is plotted versus T (the gas temperature) while pressure is kept constant, the graph is a straight line as shown below.  In practice, it is very difficult to create and control the near absolute zero condition!

Fig. 2

Since it is practically very difficult to reach very low temperatures, we have to extrapolate the graph to cross the temperature axis.  Of course, at the point of intersection, V = 0.  This means that the gas is so cold that its electrons are not spinning around the nuclei and do not need any space for their vibration.  In other words, the no need for space means Zero Volume.   -273.15oC rounded to -273oC is the extrapolated temperature that is set to be the Zero on Kelvin scale.  This makes the zero on the Celsius scale to correspond to 273 on the Kelvin scale; therefore,

K = oC + 273.

Parallel to the above discussion, if the temperatures on the graph are expressed in oF, extrapolation crosses the temperature axis at -460oF.  This constitutes the Zero on Rankin Scale.  We may write:

R = oF + 460.

Heat:

Heat is a form of energy that transfers due to a temperature difference.

Units of Heat:

The familiar unit often heard is "calorie."  One calorie 1cal is the amount of heat energy that can raise the temperature of 1gram of pure water by 1oC.  Parallel to this definition is that of kilocalorie kcal.

1 kcal is the amount of heat energy that can raise the temperature of 1kg of pure water by 1oC.   A non-Metric unit for heat energy is Btu (British thermal unit).  1 Btu is the amount of heat energy that can raise the temperature of 1lbm of pure water by 1oF.  Refer to the Chart of Units.

Specific Heat (c):

Different substances take different amounts of heat energy for one unit mass of them to warm up by one degree.  For example, if you pour 1kg of pure water (1L of water, because ρwater = 1kg/L) in a light kettle and place it on a burner and also place 1kg of iron on a same heating power burner, after the same length of time, the iron piece will be much hotter than the water.  The reason is the difference in their specific heat.  Iron takes much less heat to warm up compared to water.

The specific heat (c) of a substance is the amount of heat 1kg of that substance takes to warm up by 1oC.   On this basis, the specific heat of water is 1 kcal /(kg oC).  This is because of the way kcal was defined.  The specific heat of a few elements are given below:

cwater = 1.000 kcal / (kg oC)   or,     cwater = 1.000 cal /(g oC).

ciron = 0.108 kcal / (kg oC)   or,     ciron   = 0.108 cal /(g oC).

cAl     = 0.215 kcal / (kg oC)   or,     cAl     = 0.215 cal /(g oC).

cbrass = 0.0924 kcal / (kg oC)  or,     cbrass = 0.0924cal /(g oC).

Heat calculation:

When heat is given to a pure substance or taken from it, its temperature starts changing if phase change does not start.  During a phase change (solid to liquid, liquid to solid, liquid to vapor, or vapor to liquid ), temperature remains constant.  We will look at the following two cases:

1) Heat calculation for temperature change when phase doesn't change, and

2) Heat calculation for phase change when temperature doesn't change.

Heat Calculation Due to Temperature Change:

In this case the amount of heat given or taken is obviously proportional to mass M of the object, its specific heat c, and the temperature change ΔT. The formula is therefore,

Q = McΔT.

Example 4: Calculate the amount heat that must be given to 2.14 kg of iron to warm up from 24.0oC to 88.0oC.

Solution: Q = McΔT = 2.14kg[0.108kcal /(kgoC)](88-24)oC = 14.8 kcal.

Example 5: Calculate the amount heat that must be removed from 5.00 kg of Aluminum to cool it down from 230oC to 30oC.

Solution: Q = McΔT = 5.00kg[0.215kcal /(kg oC)](30- 230)oC = -215 kcal.

Example 6: 37.0cal of heat is given to 2.00gram of water at 12.0oC.  Find its final temperature.

Solution: Q = McΔT;   ΔT = Q/Mc = 37cal /[2gr(1cal/groC)] = 18.5oC.

or,   Tf - Ti = 18.5oC   ;    Tf - 12.0oC = 18.5oC    ;   Tf  = 30.5oC.

Thermal Equilibrium:

If a few objects initially at different temperatures are brought into contact with each other (inside a thermally insulated environment like a well-insulated ice chest or thermos), after a while, they arrive at the same temperature called the "equilibrium temperature."    At this state when no further heat transfer takes place, it is said that "thermal equilibrium" is reached.  Clearly, in the process, hotter objects lose heat while colder objects gain heat.  According to the law of conservation of energy, the heat that hotter objects lose must be equal to the heat that colder objects gain.  The following equation should be written for problems that involve thermal equilibrium processes:

-(Heat loss by hotter objects) = Heat gain by colder objects.

A (-) sign is placed on the left side of the equation to make it mathematically correct.  Heat loss is negative and heat gain is positive.  The negative of heat loss is the same thing as heat gain.

Example 7:  A 65-gram piece of aluminum at 180oC is removed from a stove and placed in 45 grams of water initially at 22oC.  Find the equilibrium temperature Teq.

Solution:  Thermal equilibrium requires that

-[ heat loss by hotter objects ]   =   heat gain by colder objects.

Here, the 180oC aluminum is the hot object and the 22oC water is the cold one.   The heat loss by aluminum must be equal to the heat gain by water.  Note that the final temperature of both water and aluminum will be the same.  We may either call this either the "final temperatureTf "  or  the "equilibrium temperature, Teq."

-Mc[Tf -Ti] Al = Mc[Tf -Ti] water ;  -65(0.215)(Tf -180) = 45(1.00)(Tf -22)

or,  -14(Tf -180) = 45(Tf -22);   -14Tf +2520 = 45Tf -990;  Tf = 59oC.

Example 8:  A 225-gram piece of hot iron is removed from an electric oven at an unknown temperature.  It is known that the iron piece has been in the oven long enough so that its initial temperature can be thought as the temperature of the oven.  The iron piece is placed in 75.0 grams of water that is held by a 45.0-gram aluminum container initially at an equilibrium temperature of 25.0oC.   The final equilibrium temperature of iron, aluminum, and water becomes 41.0oC.  Find the initial temperature of iron (oven).  Assume that the whole system is thermally isolated from the surroundings.

Solution:  Thermal equilibrium requires that

-[ heat loss by hotter objects ]   =   heat gain by colder objects.

- Mc [Tf - Ti] iron =  Mc [Tf - Ti] Al +  Mc [Tf - Ti] water

- 225(0.108)(41-Ti  =  45(0.215)(41-25)  +  75(1)(41-25)

- 24.3(41-Ti ) = 1354.8      ;     41-Ti = 55.8      ;     Ti = 96.8 oC.

Heat Calculation Due to a Phase Change:

During a phase change (solid to liquid, liquid to solid, liquid to vapor, or vapor to liquid), temperature remains constant.  For example, when ice is removed from a freezer at say -25oC,  it first absorbs some heat to become ice at 0oC.   0oC is the melting temperature of ice.  At this temperature, any heat given to the ice will be consumed for melting it and not changing its temperature.  During the melting process of ice (a phase change from solid to liquid), the temperature remains constant at 0oC.  Since temperature does not change, an equation such as Q = Mc(Tf - Ti) can not be used for heat calculation.  Here the useful equation is

Q = M Lf  where Lf is called the " latent heat of fusion."

Lf is measured for different substances at their melting/freezing points or temperatures.  Typical values may be found in texts or handbooks.

Example 9:  How much heat should be removed from 250 grams of water already at its freezing point (0oC) to convert it to ice at (0oC)?   The latent heat of fusion (freezing) for water is  Lf = 80 cal/gr.

Solution:  In this problem, the heat calculation involves a phase change only.  There is no temperature change.

Q = M Lf     ;    Q = (250gr)(80 cal/gr ) = 20,000cal.

Example 10:  How much heat should be given to 250gr of ice at 0oC to convert it to water at 40.oC ?  The latent heat of freezing or melting for water is  Lf = 80 cal /gr and the specific heat of water is 1 cal/[gr oC].

Solution:  In this problem, the heat calculation involves both a phase change and a temperature change.

Q = M Lf + Mc (Tf - Ti) ;   Q = 250(80+ 250(1)(40 - 0) = 30,000cal.

Example 11:  How much heat should be given to 250gr of ice at -20oC to bring it to boil (100oC)?

Solution:  This problem has 3 steps.  A temperature change: ice from -20oC to 0oC, a phase change: ice at 0oC to water at 0oC, and another temperature change: water from 0oC to water at 100oC.  We need three constants:

cice = 0.48cal/(gr oC),    Lf = 80 cal/gr,    and   cwater = 1cal/(gr oC).

Q =   Mcice (Tf - Ti +  M Lf   +  Mcwater (Tf - Ti).

Q = 250(0.48)[0-(-20)]  +  250(80 +  250(100-0) = 47,000 cal.

Example 12:  Draw a diagram that shows temperature change vs. heat consumed for Example 11.

Solution:

Fig. 3

Example 13:  2650 cal is given to 225 grams of ice at -12.0oC.  Find the final temperature of the result.

Solution:  All of the ice may not melt.  Let's first calculate the heat necessary for the ice to warm up to 0oC ice.

Q = Mc (Tf -Ti ) =  225gr [0.480cal /(gr oC)] [0-(-12.0)]oC = 1300 cal.

The remaining heat is  2650cal - 1300cal  =  1350 cal.   Now, let's see if this heat is enough to melt the ice.

Q = MLf  = 225gr(80cal/gr) = 18000 cal  ;   No, 1350 cal is not enough to melt all of the ice.  We need to see how much of the ice does melt.

Q = MLf     ;   1350cal  =  M (80 cal / gr)   ;   M = 17 grams.

The final result is a mixture of  17 grams of  water and (225-17) grams of ice, of course both at 0oC.

Example 14:  26500 cal of heat is given to 225 grams of ice at -12.0oC.  Find the final temperature.

Solution: All of the ice may not melt.  First calculate the needed heat for the ice to warm up to 0oC-ice.

Q = Mc (Tf -Ti ) = 225gr [0.480cal /(gr oC)] [0 - (-12.0)]oC = 1300 cal.

The remaining heat is  26500cal - 1300cal  =  25200 cal.

Now, let's see if this heat is enough to melt the ice.

Q = MLf  = (225gr)(80cal /gr ) = 18000 cal  ;  yes, 25200 cal is enough to melt all of the ice.

The remaining heat is  25200cal - 18000cal  = 7200 cal.  This heat warms up water from 0oC to Tf .

Q =  Mc( Tf - Ti ) ;   7200cal  =  225gr [(1cal/(groC)] (Tf - 0oC)  ;

Tf - 0  =  7200/225 = 32oC    ;    Tf  =  32oC.

The final result is 225 grams of water at 32oC.

Chapter 12 Test Yourself 1:

1) The temperature of an object is (a) the heat the object contains  (b) a measure of how hot or cold the object is  (c) a measure of the average K.E. of the atoms and molecules of the object  (d) both b & c.   click here.

2) The regular temperature scales are (a) Kelvin and Rankin  (b) Celsius and Fahrenheit  (c) neither a nor b.

3) The absolute temperature scales are (a) Kelvin and Rankin  (b) Celsius and Fahrenheit  (c) neither a nor b.

4) The reason for using the melting point of a pure substance in the calibration of thermometers is that (a) during melting, for example, the temperature of a pure substance remains constant  (b) it can be replicated at other similar points on this planet  (c) both a & b.      click here.

5) The reason for using the boiling point of a pure substance in the calibration of thermometers is that (a) during boiling, for example, the temperature of a pure substance remains constant  (b) it can be replicated at other similar points on this planet  (c) both a & b.

6) A phase change for a pure substance means going from (a) solid phase to liquid phase or vice versa   (b) from liquid phase to vapor phase or vice versa  (c) both a & b  (d) neither a nor b.      click here.

7) The basis for calibration of the Celsius scale is the melting and freezing points of  (a) water   (b) alcohol   (c) paraffin.

8) Fahrenheit scale also uses the melting and boiling points of water as basis for calibration  (a) True   (b) False.

9) The zero of Celsius scale corresponds to (a) 32oF   (b) 0oF   (c) 50oF.      click here.

10) The 100 on the Celsius scale corresponds to (a) 120oF   (b) 180oF   (c) 212oF.

11) Conversion from oC to oF is (a) oF = (9/5)C-32  (b) oF = (9/5)C + 32    (c) oF = (9/5)C.      click here.

12) Conversion from oF to oC is (a) oC = (5/9)(F+32)   (b)oC = (5/9)(F-32)    (c) oC = (5/9)F.

13) If a temperature difference in Celsius scale is ΔC = 20o, the corresponding ΔF is (a) 68o   (b) 36o  (c) 11o.

14) If a temperature difference in Fahrenheit scale is ΔF = 72o, the corresponding ΔC is (a) 40o   (b) 80o   (c) 150o.

15) The temperature at which both scales read the same number is (a) 35o   (b) 55o   (c) -40o.      click here.

16) In Fig. 2, V1 is the volume at T1 =100oC, and V2 the volume at T2 = 50oC.  V3 is the volume at T3 = -100oC and V4 is the volume of the gas at T4 = -200oC.  The ratio of (V1-V2) / (V3-V4) as is apparent from the line segments is (a) 1/3  (b) 1/4   (c) 1/2.      click here.

17) In Fig. 2, the ratio of (T1-T2) / (T3-T4)  is (a) 1/3  (b) 1/4  (c) 1/2     click here.

18) From the above two questions, one may conclude that (a) ΔV/ΔT = constant   (b) ΔV*ΔT = constant    (c) a & b.

19) ΔV/ΔT = constant means that the variations of V with respect to T is (a) quadratic   (b) sinusoidal   (c) linear.

20) Since V varies linearly with T, it is then (a) incorrect  (b) correct  (c) partially correct to extrapolate the volume-temperature graph of Fig. 2 like a straight line to cross the temperature axis at -273C.   click here.

21) Conversion from oF to Rankin is (a) R = oF+273   (b) R = oF+460  (c) neither one.

22) Heat is a type of energy that flows due to (a) pressure difference   (b) temperature difference  (c) a & b.

23) The SI unit for heat is (a) cal   (b) Joule   (c) kcal   (d) a & c.      click here.

24) 1kcal is the amount of heat that can change the temperature of (a) 1 gram   (b) 1 lbm   (c) 1kg of water by 1oC.

25) 1Btu is the amount of heat that can change the temperature of 1 lbm of water by (a)1oF   (b) 1oC   (c) a & b.

26) 1 cal is the amount of heat that can change the temperature of 1gram of (a) iron  (b) alcohol   (c) water by 1oC.

27) Specific heat of a substance is the amount of heat that can change the temperature of 1 unit of (a) mass  (b) volume  (c) area of that substance by 1 unit of temperature.     click here.

28) It takes (a) more  (b) less amount of heat to warm up 1 kg of aluminum than 1 kg of water by 1oC.

29) The specific heat of water is (a) 1cal /(gr oC)    (a) 1kcal /(kg oC)    (a) 1Btu /(lbm oF)    (d) a, b, & c.      click here.

30) The formula for the amount of heat with no phase change is (a) Q = McΔT   (b) Q = Mc(Tf -Ti)   (c) Q = ML  (d) a & b.

31) The formula that calculates the amount of heat given or taken in a phase change is (a) Q = MLf    (b) Q = MLV    (c) a & b.      click here.

32) The amount of heat that warms up 2.5kg of aluminum at 27oC to 67oC is (a) 21.5kcal   (b) 21500cal   (c) a & b   (d) neither a nor b.

33) The amount of heat that warms up 4.0kg of aluminum at 27.0oC by 50.0oC is (a) 43.0kcal   (b) 19.8kcal   (c) a & b.

34) The amount of heat necessary to convert 5.0kg of ice at 0oC to water at 50.0oC is (a) 250kcal   (b) 650kcal  (c) neither a nor b.      click here.

35) The amount of heat necessary to convert 5.0kg of ice at - 40.0oC to water at 50.0oC is (a) 748kcal   (b) 850kcal  (c) 450kcal.

36)  The amount of heat necessary to convert 5.0kg of ice at - 40.0oC to boiling water (100oC) is (a) 848 kcal   (b) 996 kcal  (c) 750 kcal.      click here.

37) The amount of heat that 2.00kg steam at 100oC must lose to become water at 20.0oC is  (a) -160.kcal  (b) -1240kcal   (c) -320kcal.   Note that when steam at 100oC condensates to boiling water at 100o by removing heat from it, it changes phase from vapor to liquid.  The latent heat of vaporization (condensation) for water (Lv = 539 kcal / kg).  The formula for the phase change part is, of course, Q = MLv.

38) When 4.00kg of water at 25.0oC is mixed with 2.00kg of water at 75.0oC in a well-insulated container, the equilibrium temperature is (a) 41.7oC   (b) 95.0oC  (c) 50.0oC.      click here.

39) When 8.00kg of aluminum at 200C is placed in 3.00kg of water at 25.0o, knowing that heat exchange occurs between water and aluminum only, the equilibrium temperature is (a) 71.7oC   (b) 61.0oC  (c) 88.8oC.

40) 255gr hot iron is placed in 65.0gr of water at 18.5o.  The equilibrium temperature is 88.5oC.  The initial temperature of the iron piece is  (a) 254o   (b) 422oC   (c) 277oC.      click here.

Thermal Expansion:

Generally, all materials expand when they go through a temperature increase and shrink as they are cooled down.  The reason is that at higher temperatures, not only each molecule of an object oscillates faster, but also it requires a greater space due to its increased amplitude.  Water and mercury near freezing are exceptions.  Water expands as its temperature decreases from 4oC to 0oC.  This exception has its own advantages and disadvantages.  One advantage is that the density of ice becomes less than that of water causing ice to float on water.  In winter, the ice over lakes and oceans protects the underneath water from  freezing and preserves the underwater life.  One disadvantage is the cracking of cylinder heads in engines that lack antifreeze.  We may study the thermal expansion of solids, liquids, and gases separately.  In this chapter, expansion of solids and liquids will be discussed.  The expansion of gases will be studied in Chapter 14.

1. Expansion of Solids:

Solids expand in all directions.  For a wire, expansion occurs mainly in one dimension: its length.  For a plate, expansion occurs in two dimensions.  For a solid object, expansion occurs in three dimensions.  For solids, we define the "the coefficient of linear expansion."  The same coefficient will be modified and used for surface expansion as well as volume expansion.

Linear Expansion:

Experiments have shown that when a wire or slender rod is heated, the change in length ΔL is proportional to the initial length Li , the change in temperature ΔT, and a coefficient α that depends on its material.  This proportionality may be written as:

ΔL = α Li ΔT  where α is called the "the coefficient of linear thermal expansion."

Of course, ΔL = Lf - L i   and   ΔT = Tf -Ti .

Solving for α results in:  α = ΔL/(L i ΔT).  The unit for α in SI turns out to be oC-1 because ΔL and Li have units of length and cancel each other.  Write this formula with a horizontal fraction bar and verify the unit of α.

Based on this last formula, α may be stated as the change in length per unit length  ΔL/Li  per unit change in temperature ΔT or

α = ΔL /(L i ΔT).

Example 15:  The length of a copper wire is 1000.0m at 0oC and 1001.7m at 100oC  Find its coefficient of linear expansion, α.

Solution: The change in length is ΔL = 1.70m.  This change is over 1000.0m. The change in length per unit length is

ΔL/Li  = 1.7m/1000m = 1.7x10 -3   (dimensionless)

This occurs for 100oC of temperature change.  The change in length per unit length per unit change in temp. is

α = ΔL/(L i ΔT) = 1.7x10-3/100oC  = 17x10-6  [oC]-1.

The values for α for a few materials are given in Table 1.

Table 1

Coefficients of Thermal Expansion for Some Materials

 Solids ( α ) Coeff. of Linear  Expansion  oC-1 Liquids ( β ) Coeff. of Volume Expansion  oC-1 Aluminum 24x10 -6 Alcohol, ethyl 1.1x10-4 Brass 19x10 -6 Gasoline 9.5x10-4 Brick or concrete 12x10 -6 Glycerin 4.9x10-4 Copper 17x10 -6 Mercury 1.8x10-4 Glass 9.0x10 -6 Water 2.1x10-4 Pyrex 3.3x10 -6 Air and most gases at 1atm. of pressure 3.5x10-4 Ice 52x10 -6 Iron 24x10 -6

Example 16:  A copper power line is 500 miles long at -15.0oC.  Find its final length at 90.0oC.  The coefficient of linear thermal expansion for copper is 17x10-6 oC-1 Note: Although the air temperature does not reach 90.0oC in summera copper wire under sunlight can reach that temperature.

Solution: ΔL = α Li ΔT = 17x10-6oC-1(500mi)[90.0-(-15.0)]oC = 0.85 mi.

The final length is 501mi.

 Example 17:  In a clamp-shape piece of pyrex a slender steel rod is fixed as shown. At 15oC the gap between the rod and the metal contact at B is 0.050mm.   At what temperature does the rod make contact with the metal at B?  For solution, assume the left side (Point B) to be fixed. Solution:  Both the steel rod and the pyrex clamp expand, but by different amounts.  As the rod expands to the left, the 25.45mm-segment of pyrex shifts it to the right at a slower pace because αpyrex < αsteel .     ΔL = α L i ΔT ;  0.050mm = (24-3.3)10-6 (25.45mm)(Tf -15oC)  Tf = 110oC. Fig. 4

Surface Expansion:

Again, the change in surface area ΔA is proportional to the initial area Ai ,  a constant that we may call " the coefficient of surface expansion",  and the change in temperature ΔT.  It can be shown that the coefficient of surface expansion is twice that of linear expansion or simply .  We may write:

ΔA =  Ai (2α)ΔT.

 Proof: For a square sheet of length Li that is warmed up by ΔT, each side of it gets a final length of Lf such that ΔL = α L i ΔT   or,   Lf - Li = α L i ΔT  or, Lf = Li + α L i ΔT  or, Lf  =  Li ( 1 + α ΔT ). The initial area is Li2 and the final area becomes  Lf 2.  But, Lf 2 is:  Lf 2 =  Li 2 (1 + αΔT)2  or,  Lf 2 = Li2(1+ 2α ΔT + α2ΔT2). Fig. 5  since α  is small, α2 is much smaller and can be neglected; therefore, Lf2 becomes: Lf2 = Li2 (1+ 2αΔT) or,  Af = Ai (1+2α ΔT )  or, Af -Ai = 2α Ai ΔT  or,   ΔA  = Ai (2α)ΔT.   Therefore, the coeff. of area expansion is twice the coeff. of linear expansion, α.

Example 18:  Calculate the change in the area of a 25m X 62m aluminum roof when its temperature changes from -3.0oC and 57.0oC.

Solution: ΔA = 2α Ai ΔT = 2(24x10-6 oC-1)(25m)(62m)[57.0 -(-3.0)]oC = 4.5m2.

Volume Expansion of Solids:

Once more, the volume expansion, ΔV of a solid is proportional to its initial volume Vi , the change in temperature ΔT, and a constant that we may name coefficient of volume expansion. This constant proves to be 3α ; therefore,

ΔV = Vi (3α)ΔT.

Example 19:  Calculate the volume expansion for a chunk of steel that has a volume of 35,000cm3 and its temperature changes from -15oC to 85oC.

Solution: ΔV = 3αViΔT = 3(24x10-6 oC-1)(35,000cm3)[85-(-15)]oC = 250cm3.

2. Expansion of Liquids:

For liquids, we think of volume expansion only.  Again, experiments have shown that the volume expansion ΔV of a liquid is proportional to its initial volume Vi , the change in temperature ΔT, and a constant called the "coefficient of volume expansion of the liquid, β."    Typical values for β may be found in Table 1.

The formula for the expansion of liquids is therefore:  ΔV = β Vi ΔT.

Solving for β yields:     β = ΔV/(Vi ΔT).    This means that β is the change in liquid volume per unit volume per unit change in its temperature.

Example 20:  A kettle contains a gallon (3875cm3) of water at 21oC.  It is brought to boil at 99oC.  Calculate the final volume of the water in it.

Solution:  ΔV = βVi ΔT = (2.1x10-4 oC-1)(3875cm3)[99-21]oC = 63cm3.

Vf - Vi = 63cm3    ;     Vf - 3875 cm= 63cm3    ;     Vf = 3938cm3.

3. Expansion of Gases:

For gases, like liquids, we think of volume expansion only.  For gases, pressure P is a variable in addition to V and T.  This will be discussed in Chapter 14.

Chapter 12 Test Yourself 2:

1) In general, except for a few substances at certain specific temperatures, all elements expand due to temperature increase.  (a) True   (b) False      click here.

2) Thermal expansion is due to (a) increased rate of oscillations of atoms   (b) the extra space each atom requires because of increased amplitude of oscillation   (c) a & b.

3) For expansion of solids, generally two variables are considered: (a) length & temperature   (b) pressure and temperature   (c) force and temperature.      click here.

4) For linear expansion of a wire, the change in length ΔL is proportional to (a)  the length of the wire L  (b) just the change in temperature ΔT   (c) both a & b.

5) Based on the previous question, we may write: (a) ΔL = αL  (b) ΔL= αLΔT  (c) ΔL = α ΔT   where α is the proportionality factor called the "Coefficient of  Linear Thermal Expansion."  click here.

6) From Question 5, solving for α gives: (a) α = ΔL/L   (b) α = ΔL/(LΔT)   (c) α = ΔL/(ΔT).

7) Change in length per unit length is (a) ΔL    (b)1/L   (c) ΔL/L.

8) (Change in length per unit length) per unit change in temperature, may be written as (a) ΔL/ΔT      (b) 1/LΔT    (c) ΔL/L/ΔT.      click here.

9) ΔL/L/ΔT is the same thing as  ΔL/(LΔT).    (a) True    (b) False

10) The coefficient of Linear Expansion α is defined as the "change in length per unit length per unit change in temperature."  (a) True    (b) False

11) If we write ΔL as Lf - Li ,  α becomes α = (Lf - Li)/(L ΔT).   Solving for Lf yields (a) Lf = Li + LiαΔT   (b) Lf = Li [1+αΔT]   (c) both a & b.   click here.

12) The term Liα ΔT is therefore (a) the overall change in length   (b) the change in the initial length   (c) both a & b.

13) The reason we may simply use 2α  for the area expansion of a solid sheet is that, in the derivation, the term ( ΔL)2 is  (a) very small compared to other terms  (b) has a different unit   (c) both a & b.      click here.

14) The reason we use 3α for volume expansion of a solid is that, in the derivation, the terms  (ΔL)2 and (ΔL)3 are  (a) very small compared to other terms  (b) have different units   (c) both a & b.

15) For expansion of liquids, there is only one formula as opposed to the expansion of solids that have 3 formulas.  (a) True   (b) False      click here.

16) The unit for α = ΔL/(LΔT) is   (a) (temperature)-1   (b) oC-1   (c) oF-1   (d) a, b, & c.

17) To calculate ΔL for a copper power line 500 km long for a temperature change of ΔT =150oC, the equation ΔL = Liα ΔT may be used.  The result is (a) 1.28km   (b) 128m   (c) both a &b.     click here.

18) The formula that calculates the change in the volume of a liquid is (a)  ΔV = Vi ΔT    (b)  ΔV = Viβ ΔT    (c)  ΔV = β ΔT  where β is the "Coefficient of Thermal Volume Expansion."

Problem: An aluminum kettle with a capacity of 2000cm3 is full of water at 0oC.  It is heated up to 90.0oC and cooled down back to its initial temperature.  Answer the following questions:

19) If the kettle did not have water in it and it was a solid piece of aluminum in the same shape of the kettle, its volume expansion would be  (a) 3.24cm3   (b) 2.34cm3   (c) 12.96cm3.   click here.

20) The volume expansion of the kettle itself (when filled with water) is (a) equal to the volume expansion of the solid kettle   (b) less than the volume expansion of the solid kettle  (c) greater than the volume expansion of the solid kettle.

21) The volume expansion of the 2000cm3 water in the kettle is (a) 7.8cm3    (b) 37.8cm3  (c) 3.8cm3.

22) If the kettle did not expand, the volume of water that would have flown out of it due to thermal expansion would be (a) 37.8cm3  (b) 33.5cm3  (b) 30.0cm3.   click here.

23) Since in the actual case, the kettle expands anyway, the amount of water that flows out is (a) 42.1cm3  (b) 24.8cm3  (c) 30.0cm3.   click here.

Problems:

1) Convert (a) 77oF to oC  and  (b) 85oC to oF.    (c) If in a 24-hour period the difference between the coldest and the hottest temperatures in Celsius scale is ΔC=18.0o,  what is this difference in Fahrenheit scale?  (d) Convert -40oF to oC.

2) (a) How much heat must be removed from a 40.0-kg chunk of iron initially at 450.0o to cool it down to 30.0oC?   If this amount of heat is given to water initially at 25.0oC, how much water can be brought onto the verge of boiling?  ciron = 0.108 kcal/(kgoC).

3) Convert Btu to calorie by first writing down the definition of Btu and then converting each participating unit to its equivalent.  Note that 1lbm = 453.6 grams, and 1oF = (5/9)oC.

4) 257,600J of heat is given to 11.6kg of brass initially at 23.0oC.  Calculate (a) its final temperature.  1kcal = 4186J and cbrass = 0.0924 kcal/(kgoC).   (b) If the same amount of heat were given to the same amount of water at the same temperature, what would its final temperature be?

5)  2.00kg of water initially at 12.0oC is mixed with 2.00kg of water initially at at 48.0oC in a fully insulated environment such that heat exchange occurs between the two only.  Find (a) the equilibrium temperature.  (b) Find the equilibrium temperature if the second water portion has a mass of 4.00kg.  (c) Is the answer to Part (a) halfway between the two temperatures?  (d) Is the answer to Part (b) 2/3 of the way from 12oC and 1/3 of the way from 48oC?  Why?  (e) If the specific heats were different (two different substances), would the case be the same?

6) 5.00kg of copper initially at 224oC is placed in 2.00kg of water initially at 28.0oC.  Find the equilibrium temperature.  ccopper = 0.0923 kcal/(kgoC).

7) 375 grams of iron initially at an unknown temperature is placed in 125 grams of water initially at 21.0oC.  The equilibrium temperature becomes 68.0oC.  Find the initial temperature of the iron.

8) 401.0 grams of brass initially at 99.9oC is placed in a calorimeter that contains 120.0grans of water at 21.0oC.  The calorimeter's inner pot is made of 70.0 grams of aluminum and of course initially has the same temperature as the water it holds.  Heat exchange occurs between brass, water, and aluminum.  An equilibrium temperature of 38.0oC is reached.  Find the specific heat of brass.

9) How much heat should be given to 500.0gr of ice at -20.oC to bring it to boil?  cice = 0.48 cal /(gr oC),  Lf = 80.0 cal/gr , and cwater = 1 cal /(gr oC).

10) A power line made of copper is 600 miles long and at -25.0oC.  Find the change in its length when the temperature is 95.0oC.  The coefficient of linear expansion of copper is 17x10 -6 oC-1 Note that in summer, although the air temperature does not reach 95.0oC; however, a copper wire under sunlight can reach that temperature.

11) A tank contains 256 gallons of water at 14.6oC.  It is brought to boil at 99.6oC.  Calculate the final volume of the water knowing that the coefficient of thermal volume expansion of water is 2.1x10 -4 oC-1.

12) In a similar manner as shown under "Surface Expansion," show that for a cube of initial length Li and final length Lf on each side, the coefficient of volume expansion is equal to 3α.