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 (phase change).   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", 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.  What 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 number?

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, volume decrease for a gas, is proportional to 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, 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: 

A familiar unit often heard is calorie.  One calorie (1 cal) is the amount of heat energy that can raise the temperature of 1 gram 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 1lb_m 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 1 kg of water ( 1 liter of water, because  ρ_water = 1 kg / L ) in a light kettle and place it on a burner and also place 1 kg of iron ( for which volume is about 1/8 of a liter ) on a similar heating strength burner, after the same length of time, the iron piece will be hotter than the water.  The reason is the difference in their specific heat.  Iron takes much less heat to become hot to touch than water.

The specific heat (c) of a substance is the amount of heat 1 kg 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:

c_water = 1.000 kcal / ( kg °C)    or        c_water = 1.000   cal / ( g °C)

c_iron   = 0.108 kcal / ( kg °C)    or        c_iron    = 0.108   cal / ( g °C)

c_Al     = 0.215 kcal / ( kg °C)    or       c_Al     = 0.215    cal / ( g °C)

c_brass = 0.0924 kcal / ( kg °C)  or        c_brass = 0.0924  cal / ( g °C)

Heat calculation:

If a substance is given heat to or taken heat from, its temperature changes if its phase doesn't change.  During a phase change (solid to liquid, liquid to solid, liquid to vapor, or vapor to liquid ), temperature remains constant.  We may look at two cases when doing heat calculations.  Case 1, when heat is given or taken, and temperature change occurs only.  Case 2, when heat is given or taken, and phase change occurs only.

Heat Calculation When Temperature change Occurs Only:

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 24.0°C to 88.0°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

                                     T_f - T_i = 18.5°C    ;     T_f - 12.0°C = 18.5°C     ;    T_f  = 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 (T_eq).

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 T_f or T_eq, the equilibrium temperature.

- Mc [T_f - T_i] _Al =  Mc[T_f - T_i] _water    ;    - (65)(0.215)(T_eq - 180) = (45)(1.00)(T_eq - 22)    ;

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

Example 8:  A 225-gram piece of hot iron is removed from an electric oven whose temperature is unknown.  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 gram 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 [T_f - T_i] _iron =  Mc [T_f - T_i] _Al +  Mc [T_f - T_i] _water

- (225)(0.108)(41-T_i )  =  (45)(0.215)(41-25)  +  (75)(1)(41-25)    ;    - 24.3(41-T_i ) = 1354.8  ;

41-T_i = 55.8    ;    T_i = 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 ( T_f - T_i ) can not be used for heat calculation.  Here the useful equation is  Q = M L_f  where L_f is called the " latent heat of fusion. "  L_f 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  L_f = 80. cal/gr.

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

Q = M L_f     ;    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  L_f = 80. cal/gr and the specific heat of water is 1.000 cal/[gr °C].

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

Q = M L_f  + Mc (T_f - T_i)    ;    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 100.°C).  We need three constants: c_ice = 0.48 cal / (gr °C),    L_f = 80. cal/gr ,     and c_water = 1.0 cal / (gr °C).

Q = Mc_ice (T_f - T_i)  +  M L_f   +  Mc_water (T_f - T_i)

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 ( T_f - T_i )  =  (225gr)(0.48 cal/gr °C)[ 0.0 - (-12.0)]°C =  1296 cal.

The remaining heat is  2650cal - 1296cal  =  1354 cal.   Now, let's see if this heat is enough to melt the ice.

                Q = ML_f  = (225gr)(80. cal /gr ) = 18000 cal    ; No, 1354 cal is not enough to melt all of the ice.  We need to see how much of the ice does melt. 

                Q = ML_f     ;    1354cal  =  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.

Chapter 12 Test Yourself 1:

1) 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) 0.0ºF   (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 Metric unit for heat is (a) calorie (cal).    (b) kilocalorie (kcal).   (c) a & b.      click here

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

25) 1Btu is the amount of heat that can change the temperature of 1 lb_m 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.

29) The specific of water is (a) 1cal /(gr ºC)    (a) 1kcal /(kgr ºC)    (a) 1Btu /(lb_m 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(T_f -T_i)   (c) Q = ML  (d) a & b.

31) The formula for the amount of heat with phase change is (a) Q = ML_f .   (b) Q = ML_i.    (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 bring 5.0kg of ice at 0.0ºC to water at 50.0ºC is (a) 250.kcal.   (b) 650.kcal.  (c) neither one.      click here

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

36)  The amount of heat necessary to bring 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 100.0ºC must lose to become water at 20.0ºC is  (a) -160.kcal  (b) -1240kcal.   (c) -320kcal.   Note that when steam at 100ºC condensates to become boiling water at 100ºC by heat removal from it, it changes phase from vapor to liquid.  The latent heat of vaporization (condensation) for water (L_v = 539 kcal / kg).  The formula for the phase change part is, of course, Q = ML_v.

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 place 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 their temperature increases.  The reason is that at higher temperatures, not only molecules of an object oscillate faster, but also they require greater spaces due to their increased amplitudes.  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 more than that of water causing ice to float on water.  In winter, 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 and volume occupying object, expansion occurs in three dimensions.  For solids, we define the " thermal coefficient of linear expansion ."   The same coefficient will be modified and used for surface expansion as well as volume expansion.

Linear Expansion:

Experiments show that when a wire or slender rod is heated, the change in length (ΔL) that it gains 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 L_i ΔT    where α is called the " thermal coefficient of linear expansion. "

Of course ΔL = L_f -  L _i   and   ΔT = T_f - T_i .

Solving for α yields:    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 is equal to the change in length per unit length (ΔL / L _i )  per unit change in temperature or  

ΔL / [ L _i ΔT ] .

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

                                                Table 1

         Thermal Coefficient of Expansion [ °C ^-1] for Some Materials

Example 16:  A copper power line is 500.00 miles long at -15°C.  Find its final length when the temperature is 90.°C.  The thermal coefficient of linear expansion of copper is 17x10 ^-6 °C^-1 .  Note that in summer, although the air temperature does not reach 90.°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. - (-15 )] °C  = 0.85 mi

                    The final length is 500.85mi

Surface Expansion

Again, the change in surface area ( ΔA ) is proportional to the change in temperature ( ΔT ), the initial area ( A _i ), and a constant we may call " the thermal coefficient of surface expansion". This constant can be shown that is equal to (2α).  We may write:

ΔA = 2α  A _i ΔT

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 A_i 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 V_i , the change in temperature ΔT, and a constant that we may name thermal coefficient of volume expansion.  This constant proves to be 3α ; therefore,

ΔV = 3α V_i 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 V_i 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 V_i , the change in temperature ΔT, and a constant that we may name "thermal 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 V_i ΔT.

Solving for β yields:     β = ΔV / (V_i ΔT).    This means that β is the change in liquid volume per unit volume per unit change in its temperature.  That's how β is defined.

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 V_i ΔT    ;    ΔV  =  ( 2.1x10 ^-4 °C^-1)( 3875cm³)[99 -21]°C = 63 cm³

  V_f - V_i = 63 cm³    ;     V_f - 3875 cm³   =  63 cm³    ;    V_f = 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 is discussed in Chapter 14.

Chapter 12 Test Yourself 2:

1) In general, except for a few substances at certain specific temperatures, all elements expand with 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 (constant) called the "Coefficient of Linear Thermal 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, ΔT 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 Thermal 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 L_f - L_i, the expression for α  becomes α =  ΔL / (L ΔT).  Solving for L_f yields (a) L_f = L_i + L_iα ΔT    (b)  L_f = L_i [ 1 + α ΔT }    (c) both a & b.      click here

12) The term L_iα 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 may simply use 3α  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.   The reason is that ΔL and L have the same unit and cancel.

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 = L_iα 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 = V_i ΔT    (b)  ΔV = V_iβ ΔT    (c)  ΔV = β ΔT    where β is the "coefficient of thermal volume expansion."

Problem: An aluminum kettle with a capacity of 2000.cm³ is full of water at 0.0°C.  It is brought 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 with water in it 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 2000.cm³ 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?  c_iron = 0.108 kcal/(kg°C).

3) Convert Btu to calorie by first writing down the definition of Btu and then replacing each participating quantity to its equivalent in the definition of the calorie.  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 c_brass = 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 and 1/3 of the way from 48?  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.  c_copper = 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?  c_ice = 0.48 cal /(gr °C),  L_f = 80.0 cal/gr , and c_water = 1.0 cal /(gr °C).

10) A power line made of copper is 600. miles long at -25.0°C.  Find the change in its length when the temperature is 95.0°C.  The thermal coefficient of linear expansion of copper is 17x10 ^-6 °C^-1 .  Note that in summer, although the air temperature does not reach 95.°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. 

Answers:

1) 25°C, 185°F, 32.4°F, -40°C     2) 1810 kcal ,  24.1 kg.    3) 252cal  or  0.252 kcal

4) 80.4°C, 28.3°    5) 30.0°C,  36.0°C, Yes, Yes, No.    6) 64.7°C    7) 213°C

8) 0.0924 kcal/(kg°C)      9) 94800cal    10) 1970m   11) 260.6 gal