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 in the American system of units, and Celsius and Kelvin in the Metric System.

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

Fahrenheit and Celsius Scales:

If an unmarked thermometer is placed in a mixture of ice and water, the mercury 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 mercury 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, 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 32° F and boils at 212° 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.

Fig. 1

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

Solution: Using the formula for C, we get:    °C = (5/9)[ F - 32 ]    ;    °C = (5/9) [ 77 - 32] = 25

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

Solution:  100°C difference on the Celsius scale corresponds to 180°F difference on the Fahrenheit scale.  Using a proportion, the difference in Fahrenheit is

ΔF / ΔC  = 180° / 100°    ;     ΔF / 24° = 9/5    ;     ΔF = 24° (1.8)    ;     ΔF = 43°F.

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 ceases and stops.  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:

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.  The temperature (-273°C) is chosen to be the Zero on the Kelvin scale.  This makes the zero on the Celsius scale to correspond to 273 on the Kelvin scale; therefore,

K = °C + 273

Parallel to the above discussion, if the temperatures on the graph are expressed in °F, extrapolation crosses the temperature axis at -460°F.  That  constitutes the basis for the "Rankin Scale."  We may write:

R = °F + 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 1°C.  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 1°C.   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 1°F.  Refer to the Chart of Units.

Specific Heat (c):

Different substances take different amounts of heat energy for one unit of mass of them to warm up by one degree.  For example, if you pour 1kg of water ( 1 liter of water, because  ρwater = 1 kg/L ) in a light kettle and place it on a burner and also place 1kg of iron (the volume of which is about 1/8 of a liter) 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 1°C.   On this basis, the specific heat of water is 1 kcal /(kg °C).  This is because of the way kcal was defined.  The specific heat of a few elements are given below:

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

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

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

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

Heat calculation:

When heat is given to a pure substance or taken from it, its temperature starts changing if a 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, and

2) Heat calculation for phase change.

Heat Calculation When Temperature changes Only (No Phase 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.C to 88.C.

Solution:  Q = McΔT ;  Q = ( 2.14kg )[0.108 kcal /(kg °C )] ( 88 - 24 )°C  = 14.8 kcal.

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

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

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

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

Tf - Ti = 18.5°C    ;     Tf - 12.0°C = 18.5°C     ;    Tf  = 30.5°C.

Thermal Equilibrium:

When a few objects at different temperatures are brought into contact with each other, after a while, they arrive at the same temperature that is called the "equilibrium temperature."    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 state at which all objects have arrived at the same temperature and no further heat transfer takes place is called "thermal equilibrium."   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 therefore a positive number and can be set equal to the heat gain.

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

Solution:  Thermal equilibrium requires that -[ heat loss by hotter objects ]   =   heat gain by colder objects.  To each side of the equation a  Q = McΔT must be applied.  Note that the final temperature of both water and aluminum will be the same.  We may either call this final temperature Tf or Teq, the equilibrium temperature.

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

-14(Teq - 180) = (45)(Teq - 22)  ; -14Teq + 2520 = 45Teq - 990  ;  3510 = 59Teq  ;   Teq = 59°C.

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 25.0°C.   The final equilibrium temperature of iron, aluminum, and water becomes 41.0°C.  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 °C.

Heat Calculation (Phase change Only):

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 -25°C,  it takes some heat to first become ice at 0°C.  0°C is the melting temperature of ice.  At this temperature, any heat given to the ice will be consumed for melting and not temperature change.  During the melting process (a phase change), the temperature remains constant at 0°C.  Since temperature does not change, an equation such as Q = M c ( 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 (0°C) to convert it to ice at (0°C)?   The latent heat of 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 = (250 gr)( 80 cal/gr ) = 20,000 cal.    (2.0x104 cal.)

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

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

Q = M Lf  + Mc (Tf - Ti)    ;    Q = (250)(80 +  (250)(1)(40 - 0)  =  30,000cal.  (3.0x104 cal.)

Example 11:  How much heat should be given to 250gr of ice at -20°C to convert it to water at boiling (100°C)?

Solution:  This problem has 3 steps.  A temperature change, ( ice from -20°C to 0°C ), a phase change  ( ice at 0°C to water at 0°C ), and another temperature change ( water from 0°C to water at 100C).  We need three constants: cice = 0.48 cal / (gr °C),    Lf = 80 cal/gr ,     and cwater = 1cal / (gr °C).

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

Q = (250)(0.48)[ 0 - (-20) ]  +  (250)( 80)  +  (250)( 100 - 0) =  47,400 cal.

2400cal                     20,000cal        25,000cal

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

Solution:

Fig. 3

Example 13:  2650 cal of heat is given to 225 grams of ice at -12.0°C.  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 0°C-ice.

Q =  Mc ( Tf - Ti )  =  (225gr)(0.480 cal/gr °C)[ 0 - (-12.0)]°C =  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)(80. cal /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 0°C.

 Example 14:  26500 cal of heat is given to 225 grams of ice at -12.0°C.  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 0°C-ice. Q =  Mc ( Tf - Ti )  =  (225gr)(0.480 cal/gr °C)[ 0.0 - (-12.0)]°C =  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)(80. cal /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 0°C to Tf .     Q =  Mc ( Tf - Ti ) ;   7200cal  =  (225gr)(1 cal/gr°C)( Tf - 0°C )  ;      Tf - 0.0°C  =  7200 / 225 =  32°C    ;    Tf  =  32°C.  The final result is 225 grams of water at 32°C.

Chapter 12 Test Yourself 1:

1) The temperature of an object is (a) the heat that object contains   (b) a measure of how hot or cold that object is  (c) a measure of the average K.E. of the atoms and molecules of that 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) The formula that converts ºC to ºF is  (a) ºF = (9/5)C - 32   (b) ºF = (9/5)C + 32    (c) ºF = (9/5)C.      click here

12) The formula that converts 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 the Celsius scale is ΔC = 20º, the corresponding ΔF is (a) 68º   (b) 36º   (c) 11º.

14) If a temperature difference in the Fahrenheit scale is ΔF = 72º, the corresponding ΔF 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) (a) 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 the variations of V with respect to T is linear, 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) The formula that converts Fahrenheit temperature to Rankin is (a) R = °F + 273   (b) R = °F + 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) 1ºR   (c) a & b.

26) 1 cal is the amount of heat that can change the temperature of 1 gram 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 /(kgr 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.

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) 250.kcal   (b) 650.kcal  (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) 450.kcal.

36)  The amount of heat necessary to convert 5.0kg of ice at - 40.0ºC to boiling water 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 100ºC 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 that absorbs no heat, the equilibrium temperature is (a) 41.7ºC   (b) 95.0ºC  (c) 50.0ºC.      click here

39) When 8.000kg of aluminum at 200.0ºC is placed in 3.000kg of water at 25.0ºC, 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 18.5ºC.  The equilibrium temperature is 88.5ºC.  The initial temperature of the iron piece is  (a) 254ºC   (b) 422ºC   (c) 277ºC.      click here

Thermal Expansion:

Generally, all materials, except a few, expand when they go through a temperature increase.  The reason is that at a higher temperature, not only each molecule of an object oscillates faster, but also it requires a greater space due to its increased amplitude.  As an exception, water, for example, expands when its temperature decreases from 4°C to 0°C.  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 them from further freezing and preserves 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 or a slender rod, expansion occurs mainly in one dimension.  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 its initial length L i , 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 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 is °C -1 because ΔL and L i have units of length and cancel each other.  Write this formula with a horizontal fraction bar and verify the unit for 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

ΔL / (L i ΔT).

 Example 15:     A copper wire is 1000.0m  at 0°C and 1001.7m at 100°C.  Find its coefficient of linear expansion, α. Solution: The change in length is ΔL = 1.70m.  This change is over 1000m, the change in length per unit length is  ΔL / L i  = 1.7m / 1000m = 1.7x10 -3 dimensionless.  This occurs for 100°C of temperature change.  The change in length per unit length per unit change in temp. is  α = ΔL /( L i  ΔT ) = (1.7x10 -3 ) /100°C  = 17x10 -6 °C-1.

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

Table 1

Coefficients of Thermal Expansion for Some Materials

 Material (Solid) ( α ) Coeff. of Linear  Expansion °C-1 Material (Liquid) ( β ) Coeff. of Volume Expansion °C-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 power line made of copper is 500.00 miles long at -15.0°C.  Find its final length when the temperature is 90.0°C.  The coefficient of linear expansion for copper is 17x10 -6 °C-1 Note that in summer, although the air temperature does not reach 90.0°C ( 194°F ); however, a copper wire under sunlight can reach that temperature.

Solution:   ΔL = α L i ΔT   ; ΔL = ( 17x10-6 °C-1)(500.00mi )[ 90.0 - (-15.0 )] °C  = 0.85 mi.

The final length is 500.85mi.

 Example 17:  In a clamp-shape piece of pyrex a slender steel rod is fixed as shown. At 15°C 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 (25.4mm)(Tf - 15°C) ;  Tf = 110°C. Fig. 4

Surface Expansion

Again, the change in surface area ( ΔA ) is proportional to the initial area ( A i ),  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.

 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 + α ΔT )2 or,  Lf 2 =  Li 2( 1 + 2α ΔT  + α 2ΔT2). Fig. 5  since α  is small, α 2 is much smaller and can be neglected; therefore, Lf 2 becomes: Lf 2 =  Li 2( 1 + 2α ΔT )  or,   Af = Ai ( 1 + 2α ΔT )  or, Af -Ai = 2α Ai ΔT      or,      ΔA = 2α Ai ΔT.   Therefore, the coeff. of area expansion is twice the coeff. of linear expansion, α.

Example 18:  Calculate the change in the area of an aluminum roof (25m x 62m) when its temperature changes from -3.0°C and 57.0°C.

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

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 -15°C to 85°C.

Solution:  ΔV = 3α Vi ΔT    ;    ΔV  =  3( 24x10-6 °C-1)( 35,000cm3)[85 -(-15)]°C = 250 cm3

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 thermal 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 (3875 cm3) of water at 21°C.  It is brought to boil at 99°C.  Calculate the final volume of the water in it.

Solution:  ΔV =  β Vi ΔT    ;  ΔV  =  ( 2.1x10 -4 °C-1)( 3875cm3)[99 -21]°C = 63 cm3

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

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) just 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 = α L     (b) ΔL = α L ΔT        (c) ΔL = α ΔT    where α is the proportionality factor called the "Coefficient of Linear Expansion."      click here

6) From the previous question, solving for α  gives: (a) α = ΔL / L   (b) α =  ΔL / (L ΔT)   (c) α = ΔL / (ΔT).

7) Change in length per unit length may be written as (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 / (L ΔT).    (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, the expression for α  becomes  α =  (Lf - Li) / (L ΔT).   Solving for Lf yields (a) Lf = Li + Liα ΔT    (b)  Lf = Li [ 1 + α ΔT }    (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  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 may simply use  for the  volume expansion of a solid pure substance 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 / (L ΔT) is (a) (temperature unit)-1   (b) °C-1   (c) °F-1   (d) a, b, & c.

17) To calculate the change in length of a copper power line 500.km long for a temperature change of ΔT =150°C, 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 volume of a liquid is (a)  ΔV = Vi ΔT    (b)  ΔV = Viβ ΔT    (c)  ΔV = β ΔT  where β is the "coefficient of volume expansion."

Problem: An aluminum kettle with a capacity of 2000cm3 is full of water at 0°C.  It is heated up to 90.0°C 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 due to expansion is (a) 42.1cm3    (b) 24.8cm3    (c) 30.0cm3.      click here

Problems:

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

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

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

4) 257,600J of heat is given to 11.6kg of brass initially at 23.0°C.  Calculate (a) its final temperature.  1kcal = 4186J and cbrass = 0.0924 kcal/(kg°C).  (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 12.0°C is mixed with 2.00kg of water initially at at 48.0°C 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 12°C and 1/3 of the way from 48°C?  Why?  (e) If the specific heats were different (two different substances), would the same be the case?

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

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

8) 401.0 grams of brass initially at 99.9°C is placed in a calorimeter that contains 120.0grans of water at 21.0°C.  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 38.0°C is reached.  Find the specific heat of brass.

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

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

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