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Physics 111

Lecture 33. Chapter 11: Heat Specific Heat. Monday, November 16, 1998. Physics 111. Exam # 3. Friday, November 20, 1998 in class Chapters 8, 9 (S. 1-8), 10 (S. 1-2, 4-6), and 11 (S. 1-5, 7). Hint : Be able to do the homework (both the

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Physics 111

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  1. Lecture 33 Chapter 11: Heat Specific Heat Monday, November 16, 1998 Physics 111

  2. Exam # 3 Friday, November 20, 1998 in class Chapters 8, 9 (S. 1-8), 10 (S. 1-2, 4-6), and 11 (S. 1-5, 7) Hint: Be able to do the homework (both the problems to turn in AND the recommended ones) you’ll do fine on the exam! You may bring one 3”X5” index card (hand-written on both sides), a pencil or pen, and a scientific calculator with you. I will put any constants, math, and Ch. 1 - 7 formulas which you might need on a single page attached to the back of the exam.

  3. The slope of this graph is given by the mass of the substance times the specific heat of the substance: Q DT Specific Heat Where Q is the heat added (lost) and DT is the temperature change, m the mass, and c the specific heat.

  4. Units!

  5. Specific Heat Specifically, the specific heat of a substance is the amount of heat required to raise the temperature of 1 kg of the substance by 1oC. Therefore, the specific heat of water, which requires 4.186 J to raise 1 g by 1oC, is 4186 J/kg/oC. Water has a very HIGH specific heat compared to most other substances.

  6. The HEAT is on! We can rewrite our last equation solving for heat By convention, heat flows into a system (noted by temperature increases) are usually positive while heat flows out of a system (noted by temperature decreases) are negative. Note: we did NOT use this in lab last night.

  7. If the summer solstice occurs on June 21st, and on that day, the sun is most directly over head in the Northern Hemisphere, why is it that July and August are hotter months than June? ? The heat supplied from the Sun goes into warming up the atmosphere and the water (67% of the planet’s surface). Recall, water has a very high specific heat. Therefore, it takes quite a while for the water to warm up. Notice that in the summertime, it’s usually a couple of degrees cooler down by Lake Michigan than here in Valpo. Same physics!

  8. Calorimetry The study of the transfer of thermal energy between two or more materials in a system. We performed such an experiment in lab last week, looking the temperature change of the bath and melt water from the ice cube. The basic technique to solving calorimetry problems: Heat lost by substance 1 Heat gain by substance 2 =

  9. When 100 g of aluminum shot is heated to 100oC and placed in a 500 g water bath initially at 18.3oC, the final equilibrium temperature of the mixture is 21.7oC. What is the specific heat of aluminum? ? Let’s first calculate the heat gained by the water:

  10. When 100 g of aluminum shot is heated to 100oC and placed in a 500 g water bath initially at 18.3oC, the final equilibrium temperature of the mixture is 21.7oC. What is the specific heat of aluminum? ? Now, let’s look at the heat lost by the aluminum (which must equal the gain by the water:

  11. d solid liquid gas Latent Heat As we also saw in our lab experiment last week, when a substance changes phase, extra energy is either released or required. We classify matter as existing in one of three phases:

  12. T Latent Heat If we start out with our substance in a gaseous phase, then how do we get it to become a liquid? Think of the result of a lower temperature in terms of the molecular argument about the speed of the molecules... Cool It! Molecules are more likely to stick together. vrms Energy is released as a gas becomes a liquid!

  13. Latent Heat So, in order to get a substance to change phase from a liquid to a gas, we must... Heat It! We have to add energy to vaporize a liquid. The amount of energy we must add to get a substance to change from a liquid to a gas is the same as the amount of energy that is released when the same substance changes phase from a gas to a liquid…and is known as Lv the latent heat of vaporization

  14. T Latent Heat SIMILARLY... If we start out with our substance in a liquid phase, then how do we get it to become a solid? Again, think of the result of a lower temperature in terms of the molecular argument about the speed of the molecules... Cool It! Molecules are more likely to stick together. vrms Energy is released as a liquid becomes a solid!

  15. Latent Heat So, in order to get a substance to change phase from a solid to a liquid, we must... Heat It! We have to add energy to melt a solid. The amount of energy we must add to get a substance to change from a solid to a liquid is the same as the amount of energy that is released when the same substance changes phase from a liquid to a solid…and is known as Lf the latent heat of fusion

  16. Latent Heat On what property of our substance could the latent heats (that is, the amount of heat necessary to melt our substance from a solid to a liquid of to vaporize our liquid to a gas) possibly depend? How much of the substance we’ve got. Clearly, the more of it I’ve got, the more heat I’m going to have to add to get a given temperature change. We specify this quantity with mass (m).

  17. Qp m The slope of this line is the Latent heat (L) of either fusion of vaporization, depending upon the phase change we’re examining. YES!! Latent Heat Should this graph pass through the origin? slope = L That is, how much heat do I have to add to get the phase to change for an amount m of the substance?

  18. A 100.0 g ice cube is initially at a temperature of -30.0oC. The ice cube is placed in a sealed container over a heat source. Heat is added until all of the ice has become steam at a final temperature of 120.0oC. How much total heat was added to the ice cube? ? Use the following information: cH2O(s) = 2090 J/kg/oC Lf = 3.33 X 105 J/kg cH2O(l) = 4190 J/kg/oC Lv = 2.26X106 J/kg cH2O(g) = 2010 J/kg/oC

  19. A 100.0 g ice cube is initially at a temperature of -30.0oC. The ice cube is placed in a sealed container over a heat source. Heat is added until all of the ice has become steam at a final temperature of 120.0oC. How much total heat was added to the ice cube? ? First, we must warm the ice to the temperature at which it melts (0oC). This requires heat:

  20. A 100.0 g ice cube is initially at a temperature of -30.0oC. The ice cube is placed in a sealed container over a heat source. Heat is added until all of the ice has become steam at a final temperature of 120.0oC. How much total heat was added to the ice cube? ? Next, we must melt the ice. This requires heat:

  21. A 100.0 g ice cube is initially at a temperature of -30.0oC. The ice cube is placed in a sealed container over a heat source. Heat is added until all of the ice has become steam at a final temperature of 120.0oC. How much total heat was added to the ice cube? ? Then, we must warm the water to the temperature at which it boils (100oC). This requires heat:

  22. A 100.0 g ice cube is initially at a temperature of -30.0oC. The ice cube is placed in a sealed container over a heat source. Heat is added until all of the ice has become steam at a final temperature of 120.0oC. How much total heat was added to the ice cube? ? Next, we must change the liquid water into steam. This requires heat:

  23. A 100.0 g ice cube is initially at a temperature of -30.0oC. The ice cube is placed in a sealed container over a heat source. Heat is added until all of the ice has become steam at a final temperature of 120.0oC. How much total heat was added to the ice cube? ? Then, we must warm the steam to the temperature of 120.0oC.This requires heat:

  24. A 100.0 g ice cube is initially at a temperature of -30.0oC. The ice cube is placed in a sealed container over a heat source. Heat is added until all of the ice has become steam at a final temperature of 120.0oC. How much total heat was added to the ice cube? ? The total amount of heat added is then just the sum of the heats added at each step:

  25. 2) Heat gains (losses) can only be found over intervals where no phase changes occur. Calorimetry Problems Tips for solving such problems: 1) Make sure your units are consistent!

  26. 3) Only use the latent heat formula when phase changes occur! Calorimetry Problems Tips for solving such problems: 4) Set heat loss = heat gain and use ONLY positive values of DT !!

  27. Thermal Conductivity One of three ways to transfer thermal energy from one body to another, this method is the most direct. It actually involves contact between the two substances. Recall during lab last week, after you poured the hot water into the aluminum container, how did that container feel to your hand? HOT!! Why?

  28. kth thermal conductivity Thermal Conductivity The heat from the water inside the aluminum container was conducted through the container to your hand! The ability of a material to conduct heat is yet another of the myriad of tabulated properties of materials. We’ll denote this property by the constant

  29. Thermal Conductivity How long after you poured the hot water into the container did it take for you to notice that the container was now hot? Probably not very long... How would things change if the aluminum container were thicker? Would have taken longer to notice.

  30. Units! Thermal Conductivity We call the rate at which heat is conducted across a given material the heat transfer rate(H) and define this quantity as:

  31. T2 T1 l Thermal Conductivity So, we’ve already deduced that H will go down as the thickness of the material goes up. What would H be in the following case: T1 = T2 0!

  32. T2 T1 l Thermal Conductivity Clearly, the bigger the temperature difference across the material (known as the temperature gradient), the bigger H will be. So H must be proportional to (T2 - T1). Could H depend upon anything else?

  33. T1 T2 T2 T1 l l Thermal Conductivity Well, clearly the left plate will conduct heat across itself faster than will the right plate, given the same initial temperature gradients. So H is proportional to the surface area, too.

  34. Thermal Conductivity Hence, our relationship for the heat transfer rate is given by: A the cross-sectional area T2 - T1 the temperature gradient l the thickness of the material kTH the thermal conductivity

  35. Thermal Conductivity It’s useful to note: Builder’s refer to the R-value of a material to describe the thermal insulation properties. The R-value is related to the material’s thermal conductivity by

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