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Thermal Physics. Energy in Thermal Processes. Energy Transfer. When two objects of different temperatures are placed in thermal contact, the temperature of the warmer decreases and the temperature of the cooler increases The energy exchange ceases when the objects reach thermal equilibrium

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thermal physics

Thermal Physics

Energy in Thermal Processes

energy transfer
Energy Transfer
  • When two objects of different temperatures are placed in thermal contact, the temperature of the warmer decreases and the temperature of the cooler increases
  • The energy exchange ceases when the objects reach thermal equilibrium
  • The concept of energy was broadened from just mechanical to include internal
    • Made Conservation of Energy a universal law of nature
heat compared to internal energy
Heat Compared to Internal Energy
  • Important to distinguish between them
    • They are not interchangeable
  • They mean very different things when used in physics
internal energy
Internal Energy
  • Internal Energy, U, is the energy associated with the microscopic components of the system
    • Includes kinetic and potential energy associated with the random translational, rotational and vibrational motion of the atoms or molecules
    • Also includes any potential energy bonding the particles together
thermal energy
Thermal Energy
  • Thermal Energy is the portion of the Internal Energy, U, that is associated with the motion of the microscopic components of the system.
  • Heat is the transfer of energy between a system and its environment because of a temperature difference between them
    • The symbol Q is used to represent the amount of energy transferred by heat between a system and its environment
units of heat
Units of Heat
  • Calorie
    • An historical unit, before the connection between thermodynamics and mechanics was recognized
    • A calorie is the amount of energy necessary to raise the temperature of 1 g of water from 14.5° C to 15.5° C .
      • A Calorie (food calorie) is 1000 cal
units of heat cont
Units of Heat, cont.
  • US Customary Unit – BTU
  • BTU stands for British Thermal Unit
    • A BTU is the amount of energy necessary to raise the temperature of 1 lb of water from 63° F to 64° F
  • 1 cal = 4.186 J
    • This is called the Mechanical Equivalent of Heat
problem working off breakfast
Problem: Working Off Breakfast
  • A student eats breakfast consisting of two bowls of cereal and milk, containing a total of 3.20 x 102 Calories of energy. He wishes to do an equivalent amount of work in the gymnasium by doing curls with a 25 kg barbell. How many times must he raise the weight to expend that much energy? Assume that he raises it through a vertical displacement of 0.4 m each time, the distance from his lap to his upper chest.


problem working off breakfast1
Problem: Working Off Breakfast
  • Convert his breakfast Calories, E, to joules:
problem working off breakfast2
Problem: Working Off Breakfast
  • Use the work-energy theorem to find the work necessary to lift the barbell up to its maximum height.
  • The student must expend the same amount of energy lowering the barbell, making 2mgh per repetition. Multiply this amount by n repetitions and set it equal to the food energy E:
problem working off breakfast3
Problem: Working Off Breakfast
  • Solve for n,substituting the food energy for E:
james prescott joule
James Prescott Joule
  • 1818 – 1889
  • British physicist
  • Conservation of Energy
  • Relationship between heat and other forms of energy transfer
specific heat
Specific Heat
  • Every substance requires a unique amount of energy per unit mass to change the temperature of that substance by 1° C
  • The specific heat, c, of a substance is a measure of this amount
units of specific heat
Units of Specific Heat
  • SI units
    • J / kg °C
  • Historical units
    • cal / g °C
heat and specific heat
Heat and Specific Heat
  • Q = m c ΔT
  • ΔT is always the final temperature minus the initial temperature
  • When the temperature increases, ΔT and ΔQ are considered to be positive and energy flows into the system
  • When the temperature decreases, ΔT and ΔQ are considered to be negative and energy flows out of the system
a consequence of different specific heats
A Consequence of Different Specific Heats
  • Water has a high specific heat compared to land
  • On a hot day, the air above the land warms faster
  • The warmer air flows upward and cooler air moves toward the beach
  • One technique for determining the specific heat of a substance
  • A calorimeter is a vessel that is a good insulator which allows a thermal equilibrium to be achieved between substances without any energy loss to the environment
  • Analysis performed using a calorimeter
  • Conservation of energy applies to the isolated system
  • The energy that leaves the warmer substance equals the energy that enters the water
    • Qcold = -Qhot
    • Negative sign keeps consistency in the sign convention of ΔT
calorimetry with more than two materials
Calorimetry with More Than Two Materials
  • In some cases it may be difficult to determine which materials gain heat and which materials lose heat
  • You can start with SQ = 0
    • Each Q = m c DT
    • Use Tf – Ti
    • You don’t have to determine before using the equation which materials will gain or lose heat
phase changes
Phase Changes
  • A phase change occurs when the physical characteristics of the substance change from one form to another
  • Common phases changes are
    • Solid to liquid – melting
    • Liquid to gas – boiling
  • Phases changes involve a change in the internal energy, but no change in temperature
latent heat
Latent Heat
  • During a phase change, the amount of heat is given as
    • Q = ±m L
  • L is the latent heat of the substance
    • Latent means hidden
    • L depends on the substance and the nature of the phase change
  • Choose a positive sign if you are adding energy to the system and a negative sign if energy is being removed from the system
latent heat cont
Latent Heat, cont.
  • SI units of latent heat are J / kg
  • Latent heat of fusion, Lf, is used for melting or freezing
  • Latent heat of vaporization, Lv, is used for boiling or condensing
  • Table 11.2 gives the latent heats for various substances
problem boiling liquid helium
Problem: Boiling Liquid Helium
  • Liquid helium has a very low boiling point, 4.2 K, as well as low latent heat of vaporization, 2.09 x 104 J/kg. If energy is transferred to a container of liquid helium at the boiling point from an immersed electric heater at a rate of 10 W, how long does it take to boil away 2 kg of the liquid?
problem boiling liquid helium1
Problem: Boiling Liquid Helium
  • Find the energy needed to vaporize 2 kg of liquid helium at its boiling point:
  • Divide this result by the power to find the time:
  • Some substances will go directly from solid to gaseous phase
    • Without passing through the liquid phase
  • This process is called sublimation
    • There will be a latent heat of sublimation associated with this phase change
warming ice
Warming Ice
  • Start with one gram of ice at –30.0º C
  • During A, the temperature of the ice changes from –30.0º C to 0º C
  • Use Q = m c ΔT
  • Will add 62.7 J of energy
melting ice
Melting Ice
  • Once at 0º C, the phase change (melting) starts
  • The temperature stays the same although energy is still being added
  • Use Q = m Lf
  • Needs 333 J of energy
warming water
Warming Water
  • Between 0º C and 100º C, the material is liquid and no phase changes take place
  • Energy added increases the temperature
  • Use Q = m c ΔT
  • 419 J of energy are added
boiling water
Boiling Water
  • At 100º C, a phase change occurs (boiling)
  • Temperature does not change
  • Use Q = m Lv
  • 2 260 J of energy are needed
heating steam
Heating Steam
  • After all the water is converted to steam, the steam will heat up
  • No phase change occurs
  • The added energy goes to increasing the temperature
  • Use Q = m c ΔT
  • To raise the temperature of the steam to 120°, 40.2 J of energy are needed
problem solving strategies
Problem Solving Strategies
  • Make a table
    • A column for each quantity
    • A row for each phase and/or phase change
    • Use a final column for the combination of quantities
  • Use consistent units
problem solving strategies cont
Problem Solving Strategies, cont
  • Apply Conservation of Energy
    • Transfers in energy are given as Q=mcΔT for processes with no phase changes
    • Use Q = m Lf or Q = m Lv if there is a phase change
    • In Qcold = - Qhot be careful of sign
    • ΔT is Tf – Ti
  • Solve for the unknown
your turn
Your Turn
  • You start with 250. g of ice at -10 C. How much heat is needed to raise the temperature to 0 C?
  • 10.5 kJ
  • How much more heat would be needed to melt it?
  • 83.3 kJ
your turn1
Your Turn
  • You start with 250. g of ice at -10 C. What will happen if we add 50. kJ of heat?
  • 10.5 kJ will be used to warm it up to the MP, and the rest will start melting the ice.
  • 0.119 kg will be melted
problem partial melting
Problem: Partial melting
  • A 5 kg block of ice at 0o C is added to an insulated container partially filled with 10 kg of water at 15 o C.
    • (a) Find the temperature, neglecting the heat capacity of the container.
    • (b) Find the mass of the ice that was melted.
problem partial melting1
Problem: Partial melting
  • (a) Find the equilibrium temperature.
    • First, Compute the amount of energy necessary to completely melt the ice:
problem partial melting2
Problem: Partial melting
  • Next, calculate the maximum energy that can be lost by the initial mass of liquid water without freezing it:
  • This is less than half the energy necessary to melt all the ice, so the final state of the system is a mixture of water and ice at the freezing point:
problem partial melting3
Problem: Partial melting
  • (b) Compute the mass of the ice melted.
    • Set the total available energy equal to the heat of fusion of m grams of ice, mLf:
final problem
Final Problem
  • 100. grams of hot water ( 60. C) is added to a 1.0 kg iron skillet at 500 C. What is the final temperature and state of the mixture?
final problem1
Final Problem
  • 16.7 kJ needed to warm water to BP.
  • 226 kJ needed to vaporize water
  • 199.2 kJ will be given up by skillet.
  • Final temperature will be 100. C
  • 182 kJ of heat from the skillet will be available to vaporize water
  • 81 grams of water will vaporize.