Chapter

Temperature, Heat and Phase

Blinn College - Physics 2325 - Terry Honan

. - Temperature Scales

Fahrenheit and Celsius Scales

Two commonly used temperature scales are Fahrenheit and Celsius.  In the Fahrenheit scale the temperature of the freezing point of water (at one atmosphere) is 32°F and the boiling point is 212°F.  For Celsius these two temperatures are 0°C and 100°C.  It is straightforward to convert between the two.  Consider a graph of (Subscript[T, F]) vs. (Subscript[T, C]).  We insist that these two scales are linearly related; this implies that the graph is a line.  The slope of the line is

slope=((Δ (Subscript[T, F]))/(Δ (Subscript[T, C])))=((212-32)/(100-0))=(9/5).

The intercept is 32.  It follows that the conversion is

(Subscript[T, F])=(9/5) (Subscript[T, C])+32.

Note that for temperature difference we get

Δ (Subscript[T, F])=(9/5) Δ (Subscript[T, C]).

Constant Volume Gas Thermometers

A thermometer based on the expansion of a gas at constant volume was introduced.  It was seen that a plot of pressure vs. temperature for different gases gave lines that had a common T-intercept, the temperature at zero pressure, at a very cold negative temperature.

This was the first hint of a coldest temperature that we now call absolute zero.  It should be mentioned that absolute zero is much more fundamental than just a property of cold gases.  It is a very fundamental value and the coldest temperature in all thermal experiments.  In the Celsius scale the value of absolute zero is now defined exactly as

(Subscript[T, absolute zero])=-273.15 °C.

Absolute Temperature and Kelvin

An absolute temperature scale is one that is shifted to make absolute zero, zero in that scale.  The absolute scale associated with Celsius is called the Kelvin scale.  The conversion between Celsius and Kelvin is

(Subscript[T, K])=(Subscript[T, C])+273.15 .

Temperatures in Kelvin are given as K and not °K.  We usually will take the above number to be just 273.  Note that temperature differences in Kelvin are the same as in Celsius.  We will give these Celsius temperature differences in (QuantityUnit).

Example . - Simple Conversions

Solution

(Subscript[T, C])=(Subscript[T, K])-273(QuantityUnit)=77(QuantityUnit)-273(QuantityUnit)=-196(QuantityUnit)

(Subscript[T, F])=(9/5)(Subscript[T, C])+32=-321(QuantityUnit)

Solution

(Subscript[T, F])=98.6(QuantityUnit)  and  (Subscript[T, F])=(9/5) (Subscript[T, C])+32  ⟹  (Subscript[T, C])=(5/9) ((Subscript[T, F])-32)=37(QuantityUnit)

(Subscript[T, K])=(Subscript[T, C])+273(QuantityUnit)=37(QuantityUnit) +273(QuantityUnit)=310(QuantityUnit)

. - Heat

Heat is thermal energy that flows from hot to cold, or more precisely, from higher temperature to lower temperature.  It is an essential point that heat is thermal energy that moves.  The static notion of thermal energy is a quite different thing which we will define later as internal energy.  We will use Q to denote heat.  We will choose the convention that Q is the heat added to a thermodynamic system.  When heat is removed from something we take Q to be negative.

What is the effect of heat on a system?  Suppose you add heat to a pot of water.  The heat will increase the temperature of the water, usually.  But if the water is at the boiling point the heat doesn't change the temperature; it changes the phase.  Thus, when heat is added to a system it can either change the temperature of the system or change it's phase.

Temperature Change - Specific Heat

To raise the temperature of a fixed quantity (mass) of a material by some amount Δ T requires heat that is roughly proportional to the temperature change  Q∝Δ T.  This is precise as the Δ T becomes small.  Moreover, to raise a substance by a fixed temperature requires an amount of heat proportional to m, the mass of the substance,  Q∝m.  We can combine these proportionalities and get

Q=m c Δ T.

c, the constant of proportionality is called the specific heat; it is a property of a material.  Generally, it varies somewhat with the temperature and pressure of the substance, but we will typically neglect this change as small.

The specific heat c is a property of a material.  The heat capacity C is a property of an object.

Q=C Δ T

For example, a thermos has a heat capacity; the glass in a thermos has a specific heat.  If an object is made of one material then C=m c.  If it is made of several then C=(Subscript[m, 1]) (Subscript[c, 1])+(Subscript[m, 2]) (Subscript[c, 2])+… .

Units:  The SI unit for heat is: (QuantityUnit)=joule

Alternative units for heat are (QuantityUnit) (calories), (QuantityUnit) or BTU.

1 (QuantityUnit) raises 1(QuantityUnit) of water by 1(QuantityUnit).

1 (QuantityUnit) raises 1(QuantityUnit) of water by 1(QuantityUnit).

1 BTU raises 1(Subscript[lb, m]) (pound mass) by 1F ° (Fahrenheit degree).

(BTU is awkwardly also used as a unit of power for stoves and air conditioners; in that context a BTU is implied to be BTU/h and should be written as such.)

Water has a high specific heat. Its value is:

(Subscript[c, water])=4186 (((QuantityUnit))/((QuantityUnit)·(QuantityUnit)))=1(((QuantityUnit))/((QuantityUnit)·(QuantityUnit)))=1(((QuantityUnit))/((QuantityUnit)·(QuantityUnit)))=1 (BTU/((Subscript[lb, m])F°))

Note that the specific heat of ice is different than liquid water (Subscript[c, ice])=2100(((QuantityUnit))/((QuantityUnit)·(QuantityUnit)))

Example . - Hot Coffee, Cold Mug

Solution

(Subscript[c, W])=1(((QuantityUnit))/((QuantityUnit)·(QuantityUnit))) , (Subscript[c, G])=0.20(((QuantityUnit))/((QuantityUnit)·(QuantityUnit))) ,  (Subscript[m, W])=0.25(QuantityUnit) ,  (Subscript[m, G])=0.15(QuantityUnit) ,  (Subscript[T, W])=82(QuantityUnit)   and  (Subscript[T, G])=20(QuantityUnit)

0=(Subscript[Q, tot])=(Subscript[m, W])(Subscript[c, W])(T-(Subscript[T, W]))+(Subscript[m, G])(Subscript[c, G])(T-(Subscript[T, G]))

T=(((Subscript[m, W])(Subscript[c, W])(Subscript[T, W])+(Subscript[m, G])(Subscript[c, G])(Subscript[T, G]))/((Subscript[m, W])(Subscript[c, W])+(Subscript[m, G])(Subscript[c, G])))=75.4(QuantityUnit)

Example . - Time to Boil Water

Solution

c=4186(((QuantityUnit))/((QuantityUnit)·(QuantityUnit))) ,  m=2(QuantityUnit) ,  Δ T=(Subscript[T, f])-(Subscript[T, i])=100(QuantityUnit)-20(QuantityUnit)=80(QuantityUnit)   and  𝒫=2000(QuantityUnit)

𝒫 t=m c Δ T   ⟹   t=m c Δ T/𝒫=335(QuantityUnit)=5.58(QuantityUnit)

Phase Change - Latent Heat

The phase change between solids and liquids is called fusion.  The liquid-gas transition is called vaporization.  At the temperature of a phase transition the latent heat is the amount of heat per mass needed to change the phase.

Q=±m L

The sign must be added by hand. When going from solid to liquid heat is added, so Q is positive and from liquid to solid it is negative. Similarly, from liquid to gas Q>0 and from gas to liquid Q<0.

From the table above, we can see the following.

For water: 		fusion at 0(QuantityUnit)				(Subscript[L, f])=334×(10^3)(QuantityUnit)/(QuantityUnit)
				vaporization at 100(QuantityUnit)		(Subscript[L, v])=2260×(10^3)(QuantityUnit)/(QuantityUnit)

For lead: 		fusion at 327(QuantityUnit)			(Subscript[L, f])=24.5×(10^3)(QuantityUnit)/(QuantityUnit)
				vaporization at 1750(QuantityUnit)	(Subscript[L, v])=179.5×(10^3)(QuantityUnit)/(QuantityUnit)

Note that (Subscript[L, f]) for lead is much less than for that for water. One could easily conclude, absurdly, that it is easier to melt lead than water, however it takes considerable heat to raise its temperature to 327(QuantityUnit).

Not all materials have all three phases at one atmosphere. At that pressure, helium does not have a solid phase. Also at one atmosphere, carbon dioxide lacks a liquid phase; the direct transition from solid to gas is known as sublimation and there is a latent heat of sublimation.

For (Subscript[CO, 2]): 		sublimation at -78.5(QuantityUnit)	(Subscript[L, s])=571×(10^3)(QuantityUnit)/(QuantityUnit)

Example . - Time to Boil Water (Continued)

Solution

(Subscript[L, v])=2260×(10^3)(((QuantityUnit))/((QuantityUnit))) ,  m=2(QuantityUnit)   and  𝒫=2000(QuantityUnit)

𝒫 t=m (Subscript[L, v])   ⟹   t=m (Subscript[L, v])/𝒫=2260(QuantityUnit)=37.7(QuantityUnit)

Solution

c=4186(((QuantityUnit))/((QuantityUnit)·(QuantityUnit))) ,   Δ T=80(QuantityUnit)   and  (Subscript[L, v])=2.26×(10^6)(((QuantityUnit))/((QuantityUnit)))

(Subscript[t, b])/(Subscript[t, a])=((m (Subscript[L, v])/𝒫)/(m c Δ T/𝒫   ))=((Subscript[L, v])/(c Δ T  ))=6.75

Example . - Add Ice to Water

Solution

(Subscript[m, W])=5(QuantityUnit) , (Subscript[c, I])=2100 (((QuantityUnit))/((QuantityUnit)·(QuantityUnit))) , (Subscript[c, W])=4186 (((QuantityUnit))/((QuantityUnit)·(QuantityUnit)))  and  (Subscript[L, f])=334×(10^3)(((QuantityUnit))/((QuantityUnit)))

m<(Subscript[m, 1]) ⟹ all water , (Subscript[m, 1])≤m≤(Subscript[m, 2]) ⟹ ice-water  and  m>(Subscript[m, 2]) ⟹ all ice

(Subscript[m, 1]) of ice at -25°C (Overscript[⟹, 1]) (Subscript[m, 1]) of ice at 0°C (Overscript[⟹, 2]) (Subscript[m, 1]) of water at 0°C    and   (Subscript[m, W]) of water at 20°C (Overscript[⟹, 3]) (Subscript[m, W]) of water at 0°C

0=(Subscript[Q, tot])=(Subscript[Q, 1])+(Subscript[Q, 2])+(Subscript[Q, 3])=(Subscript[m, 1])(Subscript[c, I])Δ (Subscript[T, I])+(Subscript[m, 1])(Subscript[L, f])+(Subscript[m, W])(Subscript[c, W])Δ (Subscript[T, W])

(Subscript[m, 1])=-(((Subscript[m, W])(Subscript[c, W])Δ (Subscript[T, W]))/((Subscript[c, I])Δ (Subscript[T, I])+(Subscript[L, f])))=1.08(QuantityUnit)

(Subscript[m, 2]) of ice at  -25°C (Overscript[⟹, 1]) (Subscript[m, 2]) of ice at 0°C  and   (Subscript[m, W]) of water at 20°C (Overscript[⟹, 2]) (Subscript[m, W]) of water at 0°C (Overscript[⟹, 3]) (Subscript[m, W]) of ice at 0°C

0=(Subscript[Q, tot])=(Subscript[Q, 1])+(Subscript[Q, 2])+(Subscript[Q, 3])=(Subscript[m, 2])(Subscript[c, I])Δ (Subscript[T, I])+(Subscript[m, W])(Subscript[c, W])Δ (Subscript[T, W])-(Subscript[m, W])(Subscript[L, f])

(Subscript[m, 2])=((-(Subscript[m, W])(Subscript[c, W])Δ (Subscript[T, W])+(Subscript[m, W])(Subscript[L, f]))/((Subscript[c, I])Δ (Subscript[T, I])))=39.8(QuantityUnit)

Thermal Equilibrium and the Zeroth Law of Thermodynamics

We define two systems to be in thermal equilibrium when no heat will flow between them when they are brought into thermal contact. Thermal contact means that the systems are linked in such a way where heat (and only heat) can flow from one to another. This linking could be some wall that conducts heat or it could be a conducting rod. Two systems are in thermal equilibrium if and only if they are at the same temperature.

After the three laws of thermodynamics were well established, it was realized that there was an additional assumption implicit in the notion of temperature; this became the "zeroth law." If system A is in thermal equilibrium with system B and system B is in thermal equilibrium with system C, then A is in thermal equilibrium with C. Physically, this allows us to make a thermometer. Think of system B as a thermometer. If systems A and C read the same value on B then they must be at the same temperature.

When two systems are brought into thermal contact, heat will flow from one system to the other until they reach thermal equilibrium. The time it takes to reach thermal equilibrium depends on the nature of the thermal contact. Regardless of the details of the thermal contact, the systems will eventually reach thermal equilibrium but the time required will depend on the quality of the thermal contact.

. - Phase Diagrams

Phase Diagram for Water and Other Substances

A phase diagram is a plot of pressure versus temperature showing how the phase of a substance varies with its temperature and pressure. The phase diagram above is for water and, when used interactively, shows other substances. We have discussed phases and phase transitions at 1(QuantityUnit); in the diagram, the red dashed line is 1(QuantityUnit). As the temperature at 1(QuantityUnit) increases, water changes phase from solid to liquid at 0(QuantityUnit) and from liquid to vapor at 100(QuantityUnit). The phase diagram illustrates these phase transitions at all pressures.

At a low pressure, there is a point called the triple point, where all three phases of water, or any substance, can coexist; this is at 273.16(QuantityUnit) and 0.0060(QuantityUnit) for water. Below that pressure there is no liquid phase and there is a direct sublimation phase transition from solid to gas. It should be clear from the phase diagram that the temperatures of the phase transitions vary with pressure. The phase diagram shows why water boils at a lower temperature at higher altitudes, since the pressure is lower. A pressure cooker exploits the fact that at higher pressure water boils at a higher temperature and food cooks more rapidly.

Latent heat is the energy needed to cross a phase transition curve at some pressure. Latent heats vary with pressure. The latent heat of vaporization gets smaller with increasing pressure. Surprisingly, the latent heat of vaporization goes to zero at a point known as the critical point. For water this is at 217.8(QuantityUnit) and 647.1(QuantityUnit). At temperatures and pressures higher than at the critical point there is no distinction between the liquid and gas phases; this is the super-critical region of a phase diagram.

Partial Pressure of Water and Humidity

Suppose you had a closed container with water and air and pulled all the air out to make a vacuum above the water level. Some of the water will evaporate and the region above the water would not be a perfect vacuum after the system reaches equilibrium. With the phase diagram we can see that at some temperature there is a low pressure where vapor and liquid water would be at equilibrium. This is the pressure of the vaporization curve at that temperature and it would be the small pressure above the water in the closed container.

In air at 1(QuantityUnit), this becomes the vapor pressure, or partial pressure of water, when the air and water are at equilibrium. A relative humidity of 100% corresponds to this equilibrium state. Normally the air and water are not at equilibrium and the partial pressure of water vapor is less than its equilibrium value. The relative humidity is the fraction of the partial pressure of water to its equilibrium partial pressure.

Relative Humidity=((partial pressure of water in air)/(equilibrium partial pressure of water in air))×100%