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HEAT AND ENERGY

HEAT AND ENERGY. Honors Physics March 9, 2017. Objectives. Explain heat as the energy transferred between substances that are at different temperatures. Relate heat and temperature change on the macroscopic level to particle motion on the microscopic level.

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HEAT AND ENERGY

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  1. HEAT AND ENERGY Honors Physics March 9, 2017

  2. Objectives • Explainheat as the energy transferred between substances that are at different temperatures. • Relateheat and temperature change on the macroscopic level to particle motion on the microscopic level. • Applythe principle of energy conservation to calculate changes in potential, kinetic, and internal energy.

  3. A. Defining Temperature • Temperatureis a measure of the average kinetic energy of the particles in a substance. • Addingorremoving energyusually changes temperature. • Internal energyis the energy of a substance due to both the random motions of its particles and to the potential energy that results from the distances and alignments between the particles.

  4. B. Heat and Energy • Heatis the energy transferred between objects because of a difference in their temperatures. • From a macroscopic viewpoint, energy transferred as heat tends to move from an object at highertemperature to an object at lowertemperature. • The direction in which energy travels as heat can be explained at the atomic level, as shown on the next slide.

  5. Transfer of Particles’ Kinetic Energy as Heat Energy is transferred as heat from the higher-energy particles to the lower-energy particles, as shown on the left. The net energy transferred is zero when thermal equilibrium is reached, as shown on the right.

  6. B. Heat and Energy, continued • The atoms of all objects are in continuous motion, so all objects have some internal energy. • Because temperature is a measure of that energy, all objects have some temperature. • Heat, on the other hand, is the energy transferred from one object to another because of the temperature difference between them. • When there is no temperature difference between a substance and its surroundings, no net energy is transferred as heat.

  7. B. Heat and Energy, continued • Just as other forms of energy have a symbol that identifies them (PE for potential energy, KE for kinetic energy, U for internal energy, W for work), heat is indicated by the symbol Q. • Because heat, like work, is energy in transit, all heat units can be converted to joules, the SI unit for energy.

  8. Thermal Units and Their Values in Joules

  9. C. Thermal Conduction • The type of energy transfer that is due to atoms transferring vibrations to neighboring atoms is called thermal conduction. • The rate of thermal conduction depends on the substance. • Two other mechanisms for transferring energy as heat are convection andelectromagnetic radiation.

  10. Forms of Heat Transfer

  11. D. Conservation of Energy • If changes in internal energy are taken into account along with changes in mechanical energy, the total energy is a universally conserved property. • In other words, the sum of the changes in potential, kinetic, and internal energy is equal to zero. CONSERVATION OF ENERGY PE + KE + U = 0 the change in potential energy + the change in kinetic energy + the change in internal energy = 0

  12. D. Energy Conservation

  13. D. Energy Conservation, continued • The principle of energy conservation that takes into account a system’s internal energy as well as work and heat is called the first law of thermodynamics. • The first law of thermodynamics can be expressed mathematically as follows: U = Q – W Change in system’s internal energy = energy transferred to or from system as heat – energy transferred to or from system as work

  14. EXAMPLE PROBLEMS

  15. Sample Problem #1 The First Law of Thermodynamics A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increases by 114 J during the process, what is the total amount of energy transferred as heat? Has energy been added to or removed from the refrigerant as heat?

  16. Sample Problem #1, continued Diagram: 1. Define Given: W = –135 J U = 114 J Unknown: Q = ? Tip: Work is done on the gas, so work (W) has a negative value. The internal energy increases during the process, so the change in internal energy (U) has a positive value.

  17. Sample Problem #1, continued 2. Plan Choose an equation or situation: Apply the first law of thermodynamics using the values for U and W in order to find the value for Q. U = Q – W Rearrange the equation to isolate the unknown: Q = U + W

  18. Sample Problem #1, continued 3. Calculate Substitute the values into the equation and solve: Q = 114 J + (–135 J) Q = –21 J Tip: The sign for the value of Q is negative. This indicates that energy is transferred as heat from the refrigerant.

  19. Sample Problem #1, continued 4. Evaluate Although the internal energy of the refrigerant increases under compression, more energy is added as work than can be accounted for by the increase in the internal energy. This energy is removed from the gas as heat, as indicated by the minus sign preceding the value for Q.

  20. Sample Problem #2 Conservation of Energy An arrangement similar to the one used to demonstrate energy conservation is shown in the figure. A vessel contains water. Paddles that are propelled by falling masses turn in the water. This agitation warms the water and increases its internal energy. The temperature of the water is then measured, giving an indication of the water’s internal energy increase.

  21. Sample Problem #2, continued Conservation of Energy, continued If a total mass of 11.5 kg falls 1.3 m and all of the mechanical energy is converted to internal energy, by how much will the internal energy of the water increase? (Assume no energy is transferred as heat out of the vessel to the surroundings or from the surroundings to the vessel’s interior.)

  22. Sample Problem #2, continued 1. Define Given: m = 11.5 kg h = 1.3 m g = 9.81 m/s2 Unknown: U = ?

  23. Sample Problem #2, continued 2. Plan Choose an equation or situation: Use the conservation of energy, and solve for U. PE + KE + U = 0 (PEf– PEi) + (KEf– KEi) + U = 0 U = –PEf+ PEi – KEf+ KEi Tip: Don’t forget that a change in any quantity, indicated by the symbol ∆, equals the final value minus the initial value.

  24. Sample Problem #2, continued Because the masses begin at rest, KEi equals zero. If we assume that KEfis small compared to the loss of PE, we can set KEfequal to zero also. KEf = 0 KEi = 0 Because all of the potential energy is assumed to be converted to internal energy, PEi can be set equal to mgh if PEf is set equal to zero. PEi = mghPEf = 0 Substitute each quantity into the equation for ∆U: ∆U = –PEf+ PEi – KEf+ KEi ∆U = 0 + mgh + 0 + 0 = mgh

  25. Sample Problem #2, continued 3. Calculate Substitute the values into the equation and solve: U = mgh U = (11.5 kg)(9.81 m/s2)(1.3 m) U = 1.5  102 J 4. Evaluate The answer can be estimated using rounded values. If m ≈ 10 kg and g ≈ 10 m/s2, then ∆U ≈ 130 J, which is close to the actual value calculated.

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