Thermal Properties

Ch 16 & 17: Temperature & Heat

Joseph Black (1728-1799)• English Chemist

• Founder of calorimetry technique• Proposed a theory of heat that did

not reference phlogiston, including:– The heat equation

– Specific heat capacity & Latent heat

• Widely renowned as a great teacher:

"Many were induced, by the report of his students, to attend his courses, without having any particular

relish for chemical knowledge."

Heat & Internal Energy• Heat is energy that flows from a high temperature

object to a lower temperature object– When something absorbs heat its internal energy (or the

energy of its atoms/molecules) increases– When something releases heat its internal energy decreases

• The SI units of heat are joules (J)• Two things occur when an object absorbs or releases

heat energy (Q):– The temperature will change (which is why they

expand/contract, due to changes in molecular motion)– The object (or part of it) will change phase (solid, liquid, gas)

Heat & Temperature Change

• When an solid (or liquid) of mass, m, absorbs heat and NO phase change occurs:

Q = m.c. T {Black’s Heat Equation}Q is heat absorbed/lost,

T is the temperature change

c is called the specific heat capacity

Example – Temp. Change

• When a car brakes, an amount of heat equal to 112,500 J is generated in the brake drums. If the mass of the brake drum is 28 kg and their specific heat capacity is 460.5 J kg-1 K-1, what is the change in their temperature. C7.8

5.460x28

J112500mc

Q T

T mc Q

o

Heat & Temperature ChangeSpecific Heat Capacity:

– A physical property that relates how energy absorbed reflects changes in temperature for a given substance

– In SI terms: the amount of energy required to change the temperature of 1 kg of a substance by 1oC (or 1 K)

– SI units for specific heat capacity are J kg-1K-1

– Metals tend to have low specific heat capacities (which is one reason they make great cooking vessels) See Table 2.1

– Non-metal substances tend to have higher specific heat capacities

– Water has an unusually high specific heat capacity

Heat & Temperature Change• Specific Heat Capacity:

– Metals tend to have low specific heat capacities (a reason they make great cooking vessels) See Table 2.1

– Non-metal tend to have higher specific heat capacities– Water has an unusually high specific heat capacity

• Why do substances have different sp heat cap?

• Every 1 kg of material have different numbers of molecules. If you provided the same heat to the same number of moles the temperature change would be the same!

Table 16-2Specific Heats at Atmospheric Pressures

Water 4186

Ice 2090

Steam 2010

Beryllium 1820

Air 1004

Aluminum 900

Glass 837

Silicon 703

Iron (steel) 448

Copper 387

Sliver 234

Gold 129

Lead 128

Substance Specific Heat, c [J/(kg•K)]

Heat Capacity

TCQ

The amount a given substance heats up is characterized by it’s “heat capacity”:

The concept of heat capacity is useful when a body consists of several parts of different specific heat capacities.

Heat capacity --- heat energy required to raise the temperature of a sample by 1K (=1ºC)

Depends on amount of material Units: J / KmcC

Calorimetry• A method to measure specific heat

capacity that is based on:– The principle of conservation of energy

– The known specific heat capacity for water (1000 cal/kg.oC or 4186 J/kg.oC )

• The Process:• A heated object is placed into a thermally isolated container

containing a known amount of water

Qnet = Qgained by water + Qgained by object = 0• Object and liquid reach thermal equilibrium:

Qgained by water = Qlost by object {= - Qgained by object}• The final temperature of the object/water is used to determine

the specific heat of the object

Calorimetry• A method to measure specific heat

capacity that is based on:– The principle of conservation of energy

– The known specific heat capacity for water (1000 cal/kg.oC or 4186 J/kg.oC )

• The Process:• A heated object is placed into a thermally isolated container

containing a known amount of water

Qnet = Qgained by water + Qgained by object = 0• Object and liquid reach thermal equilibrium:

Qgained by water = Qlost by object {= - Qgained by object}• The final temperature of the object/water is used to determine

the specific heat of the object

Calorimetry Problems• A piece of iron of mass 200g and temperature 300 oC is

dropped into 1.00 kg of water at 20 oC. What will be the temperature of the water when it reaches thermal equilibrium?– Use c for iron as 470 J kg-1 K-1 and for water as 4200 J Kg-1K-1

• By conservation of energy Heat lost by the iron must equal the heat gained by the water.

mironciron(300-T) = mwatercwater (T-20) 0.2 kg(470 J kg-1 K-1)(300-T) = 1kg(4200 J Kg-1K-1)(T-20)T = 26.1 oC

Example: Cooking Tips

While cooking a turkey in a microwave oven that puts out 500 W of power, you notice that the temperature probe in the turkey shows a 1°C temperature increase every 30 seconds. If you assume that the turkey has roughly the same specific heat as water (c = 4184 J/kg-K), what is your estimate for the mass of the turkey?

What Equations? Q = m.c. T ?

While cooking a turkey in a microwave oven that puts out 500 W of power, you notice that the temperature probe in the turkey shows a 1°C temperature increase every 30 seconds. If you assume that the turkey has roughly the same specific heat as water (c = 4184 J/kg-K), what is your estimate for the mass of the turkey?

kg6.3

In 30 seconds, Q = (500 W)x(30 s) = 15,000 J of heat is put out by oven

ifturkeyturkeyifturkey TTcmTTCQ

ifturkeyturkey TTc

Qm

CKkgJ

Jmturkey

1/4184

15000

Turkey Time!

Example: Cooking Tips

You place a copper ladle of mass mL= 0.15 kg (cL = 386 J/kg-K) - initially at ‘room’ temperature Tla= 20 C - into a large pot containing 0.6 kg of hot cider (cC = 4184 J/kg-K) at a temperature of 90 C. If you forget about the ladle while watching a football game on TV, roughly what is its temperature when you try to pick it up after a few minutes?

Example: Cooking TipsYou place a copper ladle of mass mL= 0.15 kg (cL = 386 J/kg-K) - initially at ‘room’ temperature Tla= 20 C - into a large pot containing 0.6 kg of hot cider (cC = 4184 J/kg-K) at a temperature of 90 C. If you forget about the ladle while watching a football game on TV, roughly what is its temperature when you try to pick it up after a few minutes?

ladlecider QQ Lofladleladlecofcidercider TTcmTTcm

ladleladlecidercider

Loladleladlecociderciderf cmcm

TcmTcmT

KkgJkgKkgJkg

KKkgJkgKKkgJkg

/38615.0/41846.0

293/38615.0363/41846.0

= 361 K = 88 C !

Phase (State) Changes & Latent Heat• When a substance absorbs/loses heat energy its temperature will

change until:– the substance reaches its “critical” temperature it will no longer change

temperature– gain/loss of additional heat energy will result in the phase transformation

of the matter from one phase to another:

Solid liquid gas .

Phase (State) Changes & Latent Heat• Why do substances change phase or states?

• Why does temperature not change during a phase change?

When the kinetic energy of the molecules become comparable to the

energy required for separation the molecules change there position

and separate (PE increase). This is a phase transition

Phase (State) Changes & Latent Heat

Solid liquid gas• The relationship that describes heat energy (Q) gained/lost

to mass of substance (m) that undergoes a phase change is:

Q = m.Lf or Q = m.Lv

L is called the specific latent heat (the SI units are J kg-1)

• Def: Specific Latent Heat is heat (energy) required to change the phase of a unit of mass (1 kg) of material at its melting or vaporization point.

• Latent heat is a physical property that describes how much energy is required to transform the mass of a substance from one phase to another (e.g. latent heat of fusion or vaporization, etc.)

Heat vs. Temperature Graph (for water)

Examples of Phase Change

Question:

So why do you feel so “cold” when you step out of the shower soaking wet?

Figure 17-22Heat Required for a Given Change in Temperature

Table 17-4Latent Heats for Various Materials

Water 33.5 x 104 22.6 x 105

Ammonia 33.2 x 104 13.7 x 105

Copper 20.7 x 104 47.3 x 105

Benzene 12.6 x 104 3.94 x 105

Ethyl alcohol 10.8 x 104 8.55 x 105

Gold 6.28 x 104 17.2 x 105

Nitrogen 2.57 x 104 2.00 x 105

Lead 2.32 x 104 8.59 x 105

Oxygen 1.39 x 104 2.13 x 105

Material Latent heat of fusion, Lf (J/kg)

Latent heat of vaporization, Lv (J/kg)

Examples: Phase Changes • An ice cube of mass 25.0 g and a temperature of -10

◦C is dropped into a bowl of punch of mass 300.0 g and temperature 20.0 ◦C. What is the temperature eventually?(Specific heat of fusion of ice = 2218 J kg-1K-1; latent heat of fusion of ice = 334 kJ kg-1.)

Examples: Phase Changes Game Plan!• Increase temp of ice from -10 to 0.

– Mct• Melt the ice cube in to water

– MLf• Increase the temperature of the former Ice cube form 0 to Tf

– mct

Thus,T = 11.9 degrees

Figure 17-14A Liquid in Equilibrium with its Vapor

Evaporation• Factors effecting evaporation rates

– Temperature– Surface Area– Vapor Pressure

• Air Flow

• Molecule concentrations

• What happens to the liquid when it evaporates?

• Is evaporation the same as boiling?

Drinking Bird

• Explain the Thermodynamic processes which occur during one cycle of the drinking bird.