Heat

Specific heat capacity

When heat flows from one material to another, the temperature of the contact layer of the cooler material increases.  From the contact layer thermal energy has to spread throughout the cooler material by conduction or convection.  How fast the temperature in the contact layer increases depends on the specific heat capacity of the material.  The specific heat capacity c is the amount of energy it takes to raise the temperature of one kg of material by 1 degree Kelvin or Celsius.

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Specific heat capacities: (kcal/(kgoC)

Water 1.0
Ice 0.49
Steam 0.48
Glass 0.20
Steel 0.11
Copper 0.092
Aluminum 0.215

In units of kcal/(kgoC) the specific heat of water is 1.  The specific heat capacity of water is approximately 4 times higher than that of air.  The exact specific heat capacity of a substance depends on the condition under which it is measured.  For gases, the specific heat capacity measured at constant volume is different from the specific heat capacity measured at constant pressure.

The smaller the specific heat capacity of a material that touches your skin, the less heat it takes to bring the temperature of the boundary layer up to the temperature of your skin.  How fast the heat is carried away from this boundary layer now depends on the thermal conductivity of the material and on whether or not convection currents are present.  To minimize heat loss from your skin, surround it by material of low specific heat capacity and low conductivity, and prevent convection.  In addition you must minimize heat loss via radiation.

Problems:

A 50g sample of copper is at 25 oC.  If 1200 J of thermal energy is added to it, what is the final temperature of the copper? 
Solution:
ΔT = ΔQ/(cm).  For copper c = 9.2*10-2 kcal/(kgoC).
ΔT = 1200 J*(1 kcal/4186 J)/(0.05 kg*9.2*10-2 kcal/(kgoC)) = 62.3 oC.
T = 87.3 oC.
An aluminum calorimeter of mass 100 g contains 250 g of water.  The calorimeter and water are in thermal equilibrium at 10 oC.  Two metallic blocks are placed in the water.  One is a 50 g piece of copper at 80oC.  The other has a mass of 70 g and is originally at a temperature of 100 oC.  The entire system stabilizes at a final temperature of 20 oC.
(a) Determine the specific heat of the unknown sample.
(b) Guess the material of the unknown sample.
Solution:
The temperature of the aluminum calorimeter and the temperature of the water are raised by 10 oC. The amount of energy gained by an object whose temperature is raised by DT is DQ = mcDT.
The water gains DQ = 0.25 kg´1 kcal/(kgoC)´10 oC = 2.5 kcal.
The aluminum gains DQ = 0.1kg´0.215 kcal/(kgoC)´10 oC = 0.215 kcal.
The temperature of the copper drops 60oC.
The copper looses DQ = 0.05 kg´0.092 kcal/(kgoC)´60 oC = 0.276 kcal.
The unknown object therefore looses DQ = (2.5+0.215-0.276) kcal = 2.439 kcal.
Its specific heat is c = DQ/(mDT) = 2.439/(0.07´80) kcal/(kgoC) = 0.436 kcal/(kgoC)
The unknown material is probably beryllium.

Radiation laws

The particles that make up an object can have ordered energy and disordered energy.  The temperature is a measure of this internal, disordered energy.  The absolute temperature of any substance is proportional to the average kinetic energy associated with the random motion of the substance.

The velocity of particles with thermal energy is changing almost all the time.  The particles are accelerating.  The atoms and molecules are themselves complicated arrangements of charged particles.  Accelerating charged particles produce electromagnetic radiation.  The power radiated is proportional to the square of the acceleration.  Higher rates of velocity change result in higher frequency (shorter wavelength) radiation.

How do we measure the intensity of radiation emitted by an object as a function of wavelength for a fixed temperature?

Problems:

Radiation carries away energy and therefore cools the object.
Some of the radiation coming from the object may be reflected or transmitted radiation.

Solution:

Use a blackbody.  A blackbody is a body that absorbs all the radiation that falls onto it.  It does not reflect any radiation.  It reaches thermal equilibrium with its surroundings, and in thermal equilibrium emits exactly as much radiation it absorbs.  It has emissivity = 1.  Emissivity measures the fraction of radiative energy that is absorbed by the body.

Experimental realization:

Use the inside of a large box (oven) held at a constant temperature T.  A small hole is cut in one side.  Any radiation that enters through the hole bounces around inside and has little chance of ever getting out again.  Eventually it gets absorbed.  Radiation coming out the hole is as good a representation of the radiation from a perfect emitter.

What is observed?

The observed intensity of the radiation emitted  as a function of wavelength can be described by the Planck Radiation Law.

The Planck Radiation Law gives the intensity of radiation emitted by a blackbody as a function of wavelength for a fixed temperature.  The Planck law gives a distribution, which peaks at some wavelength.  The peak shifts to shorter wavelengths for higher temperatures, and the area under the curve grows rapidly with increasing temperature.  The diagram below shows the intensity distribution predicted by the Plank law in J/(m2s) for blackbodies at various temperature.

The Wien law and the Stefan-Boltzmann Law can be derived from the the
Planck Radiation Law.
 

The Wien Law gives the wavelength of the peak of the radiation distribution, lmax = 3´106/T.  Here l is measured in units of nanometer = 10-9 m and T is in Kelvin.
The Stefan-Boltzmann Law gives the total energy being emitted at all wavelengths by the body.

Radiated power = emissivity ´ s ´ T4 ´ Area

Here s is the Stefan-Boltzmann constant,
s = 5.67´10-8W/(m2K4) and the temperature is measured in Kelvin.

The Wien law explains the shift of the peak to shorter wavelengths as the temperature increases, while the Stefan-Boltzmann law explains the growth in the height of the curve as the temperature increases.  This growth is very abrupt, since it varies as the fourth power of the temperature.
Links:
The Planck Law
The Wien Law and the Stefan-Boltzmann law
Black Body: The Game
The surface temperature of the sun is 5800 oC = 6073 K.  The wavelength of the peak of the distribution is 494 nanometer.  This wavelength lies in the yellow region of the visible spectrum.
In an incandescent light bulb a filament is heated to approximately 2500 oC = 2773K.  This is the maximum temperature that a tungsten filament can stand without evaporating quickly.  Compared to the sun, such a filament emits a greater fraction of its radiation in the infrared region of the electromagnetic spectrum.  The wavelength of the peak of the distribution is 1082 nanometer.  This wavelength lies in the infrared region of the spectrum.

Sunlight and light from an incandescent bulb contain all the colors of the visible spectrum.  But the intensity distribution over the different colors is different.  Sunlight appears brilliant white while a light bulb looks yellowish.

Problem:

A burning log is a black object with a surface area of 0.25 m2 and a temperature of 800 oC.  How much power does it emit as thermal radiation?

Solution:
Radiated power = emissivity ´ s ´ T4 ´ Area
= 5.67 ´ 10-8 J (s m2 K4)(1073)4(.25 m2) = 18790 J/s.
Note: The temperature is measured in Kelvin.

The emissivity of an object is the ratio of the radiative power emitted by that object at temperature T to the radiative power emitted by a blackbody of the same shape and temperature T.  It is also equal to the fraction of the incoming blackbody radiation at that temperature that is reflected by the object.  The emissivity of objects is a function of the temperature T and can be quite different for visible and infrared radiation.

The emissivity of an object for radiation emitted by high temperature sources (visible light) is easily determined by just looking at the object.  A black surface has a high-temperature emissivity near 1, while a white or shiny surface has a high-temperature emissivity near 0. 
Light-colored or reflective objects have low emissivity.  They do absorb a smaller percentage of the incoming visible radiation than do dark objects, and also emit radiation less readily.
One cannot "see" the emissivity of an object for radiation emitted by low temperature sources (infrared light).  Most materials have a low-temperature emissivity near 1, but conducting (metallic) surfaces can have a low-temperature emissivity near 0.
If you wrap a hot object in reflective aluminum foil, then the foil reflects most of the radiation emitted by the object back onto the object.  Heat loss by radiation is thus slowed down.

To reduce radiative heat transfer, use low-emissivity surfaces and allow exterior surfaces to reach the ambient temperature.

Phase Transitions

Matter exists in different states.  It can, for example, be in a solid state, a liquid state, or a gaseous state.  These states are called phases.  The atoms and molecules making up the matter move differently in different phases.

In solid the atoms vibrate about an equilibrium positions but are unable to flow or diffuse.  The intermolecular forces are very strong, and impulsive forces from collisions with thermal-energy atoms are on the average, much weaker.

In liquids the atoms or molecules can flow past each other, but they maintain close contact.  Some ordering persists over the range of a few molecular diameters.  The intermolecular forces are not negligible compared to the impulsive forces from collisions with thermal-energy atoms and molecules.

In gases the atoms or molecules move freely.  Their average spacing is much larger than the atomic or molecular diameter and the intermolecular force become negligible compared to the impulsive forces from collisions with thermal-energy atoms and molecules.  The behavior of most gases is well described by the ideal gas law, PV = NkT.
Solid Liquid Gas
 

In general, heat flows from an object of higher temperature to an object of lower temperature.  Temperature is the quantity that indicates whether or not heat will flow and in what direction it will flow.  When heat flows from a hotter to a cooler object, the temperature of the hotter object decreases, while the temperature of the cooler object rises.  The average kinetic energy of the molecules in the hotter object decreases, while the average kinetic energy of the molecules in the cooler object increases.  But when an object changes phase, its temperature does not change, even though heat is added or removed.  The melting of ice and the boiling of water are familiar examples.  During a change of phase the temperature does not change, but the internal energy does.  The internal energy is the sum of the kinetic energy of the molecules and the chemical potential energy of the molecules.  During a change of phase, the average kinetic energy of the molecules stays the same, but the average potential energy changes.
The processes represented by the red arrows require energy input.

 

The processes represented by the black arrows release energy.

 

Melting

Melting is a phase transition.  Ice and water are different phases of the same substance.  At atmospheric pressure ice exists at temperatures below 0 oC.  If we want to raise the temperature of 1 kg of ice by 1 degree Celsius, say from –5 oC to – 4 oC, we need 0.49 kcal of heat.  But if we want to melt 1 kg of ice at 0 oC, we need 80 kcal of heat.  This heat is used to break chemical bonds and is converted into potential energy.  It is called the latent heat of melting or latent heat of fusion. Latent heat is heat that changes the phase of a substance without changing its temperature.  As we add heat to a mixture of water and ice, heat will flow from the water into the ice at 0 oC until all the ice has melted.  If we could thermally isolate the mixture, it would reach thermal equilibrium at 0 oC, and no more ice would melt.  If heat is removed from water at 0 oC it freezes into ice rather tan becoming colder.  The latent heat of fusion is released.

Boiling

Boiling is a phase transition.  Water and steam are different phases of the same substance.  At atmospheric pressure water exists at temperatures between 0 oC and 100 oC.  If we want to raise the temperature of 1 kg of water by 1 degree Celsius, say from 50 oC to 51 oC, we need 1 kcal of heat.  But if we want to boil 1 kg of water at 100 oC and turn it into steam we need 540 kcal of heat.  This is called the latent heat of vaporization.  It is used to break the bonds that hold the water molecules together in the liquid.  The boiling temperature of water depends on the pressure.  The lower the pressure, the lower is the boiling temperature.  At lower pressure the molecules need less kinetic energy to escape from the liquid.

Link:  The phases of water

Problems:

How much thermal energy is required to change a 40 g ice cube from a solid at -10 oC to steam at 110 oC?
Solution:
To raise the temperature of the ice to 0 oC we need
ΔQ = 0.04 kg (0.49 kcal/(kgoC))10 oC = 0.196 kcal.
To melt the ice we need
ΔQ = 0.04 kg*80 kcal/kg = 3.2 kcal.
To raise the temperature of the water to 100 oC we need
ΔQ = 0.04 kg (1 kcal/(kgoC))100 oC=  4 kcal.
To boil the water we need
ΔQ = 0.04 kg*540 kcal/kg = 21.6 kcal.
To raise the temperature of the steam to 110oC we need
ΔQ = 0.04kg*(0.48 kcal/(kgoC))10 oC = 0.192 kcal.
The total thermal energy required is
(0.196+3.2+4+21.6+0.192)kcal = 29.188 kcal.
If 90 g of molten lead at 327.3 oC is poured into a 300 g casting form made of iron and inititially at 20 oC, what is the final temperature of the system?  Assume no energy is lost to the environment.
Solution:
The melting point of lead is 327.3oC.
Assume the final temperature of the system is T.  Then the amount of energy released by the lead as it solidifies is
DQ = 0.09 kg´(2.45´104 J/kg) = 2205 J,
and the amount of energy released as it cools is
DQ = 0.09 kg´(128 J/(kgoC))(327.3 oC - T) = (11.52 J/oC)(327.3 oC - T).
This energy is absorbed by the iron.  For the iron we therefore have
2205 J + (11.52 J/oC) (327.3oC - T) = 0.3 kg´(448 J/(kgoC))(T - 20 oC).
5975.5 J - (11.52 J/oC)T = (134 J/oC)T - 2688 J.
8663.5 J = (145.52 J/oC)T.
T = 59.5oC.

Link:

Boil water with Shockwave

Evaporation and Sublimation

The temperature of a substance is a measure of the average kinetic energy of the atoms or molecules that make up the substance.  But not all particles have the same kinetic energy, they have a distribution of energies.  Some particles in a liquid or solid may have enough kinetic energy to break the chemical bonds and leave the substance.  The liquid is evaporating and the solid is subliming.  When the particles with the highest kinetic energy leave the substance, the average kinetic energy of the remaining particles decreases.  The temperature of the substance therefore decreases.  Evaporation cools the substance.

bird.jpg (10614 bytes)

A toy that demonstrates the cooling process of evaporation is the dippy bird.  The dippy bird is made by blowing a glass tube that flows into a bulb like the neck of a funnel.  This upper bulb becomes the head of the bird.  A second bulb becomes the body.  The tube extends almost to the bottom of this lower bulb, like a straw into a softdrink.

 

bulb.gif (2604 bytes)

The bird is filled with a liquid which has a high vapor pressure.  The bird's head is covered with fuzz, which gives a large area for evaporation.  When the head is wet, evaporation causes cooling and condensation of the fluid inside the head bulb.  This causes the fluid to creep up the neck to the area of lower pressure.  The center of gravity shifts to the head end of the bird.  The bird tips into the beaker of water where the fuzz becomes wet again.  But as the bird reaches its maximum lean, the bottom end of the tube sticks out of the fluid, and the fluid can flow out of the tube back into the bottom bulb.  This shifts the center of gravity back.  The process continues as long as there is water in the beaker.

Link:

How does a dippy bird work?

Relative Humidity

Molecules can leave liquid water through evaporation, but molecules also can re-enter the liquid water if there is water vapor in the surrounding air.  We call this phenomenon condensation.  The relative humidity is the ratio of the rate of condensation to the rate of evaporation.  If twice as many molecules leave the liquid than are returning to the liquid, then the relative humidity is 0.5 or 50%.  The relative humidity depends on the temperature and on the density of the water vapor in the air.  The higher the temperature, the higher is the average kinetic energy of the molecules and therefore the rate at which they are leaving.  The higher the density of water vapor in the air, the higher is the rate the molecules are returning.  If you seal water inside a container, it will evaporate, until the rate of evaporation equals the rate of condensation.  The relative humidity is then 100% and the air is saturated.

Question:

Knoxville weather is often hot and humid in the summer.  Under those conditions, why is the time after sunrise the most comfortable time of the day?  Why is it not the time just after sunset, when the temperature is starting to drop slightly?
Answer:
Just after sunrise, when the temperature is rising, the rate of evaporation increases.  The amount of water vapor in the air, however, is still low.  The rate of condensation and the relative humidity therefore are quite low.  Sweat on the skin therefore will evaporate.  Evaporation cools the remaining sweat and, through conduction, the skin.  As the density of water vapor in the air increases, the net rate of evaporation decreases, because more and more molecules are entering the liquid.  The sweat does no longer cool the skin efficiently.  Just after sunset, when the temperature drops, the humidity increases even more, because the rate of evaporation drops, but the density of the water vapor in the air is still high.

animated.gif (50819 bytes) Would the "Dippy Bird" still work, if the humidity were 100%?

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