LECTURE 9 PHASE CHANGE. Lecture Instructor: Kazumi Tolich

Transcription

1 LECTURE 9 PHASE CHANGE Lecture Instructor: Kazumi Tolich

2 Lecture 9 2 Reading chapter Phase change and energy conservation

3 Phase changes and energy conservation 3 If heat is exchanged within a system, but not with the external world, the total energy of the system is constant. The heat flow within the system can cause changes in temperature and phases. Set the magnitude of the heat lost by one part of the system equal to the magnitude of the heat gained by another. Problem solving strategy: It is not always clear in advance whether multiple phases or only one phase is present at equilibrium. Assume one case and proceed with the calculation based on that assumption. If the result is unphysical, change the assumption and try again.

4 Solve this by applying the strategy 4 An aluminum cylinder of a mass m " = kg is removed from a liquid nitrogen bath, where it has been cooled to T ", = 196. The cylinder is immediately placed in an insulated cup containing m 1 = kg of water at T 1, = The average specific heat of aluminum over this temperature range is c " = 653 J kg 6 K. The specific heat of liquid water is c 1 = 4186 J kg 6 K. The latent heat of fusion for water is L 1, < = > J kg. What is the equilibrium temperature of this system, T <?

5 Quiz: 1 5 An aluminum cylinder is removed from a liquid nitrogen bath (below freezing point of water). The cylinder is immediately placed in an insulated cup containing water. Without doing any calculation, you do not know the final state. Which of the following is/are not possible? A. The final temperature could be higher than 0 without forming any ice. B. The final temperature could be higher than 0 with some water freezing. C. The final temperature could be 0 without forming any ice. D. The final temperature could be 0 with forming some ice. E. The final temperature could be 0 with all of water freezing. F. The final temperature could be below 0 without forming any ice. G. The final temperature could be below 0 with some water freezing. H. The final temperature could be below 0 with all of water freezing.

6 Quiz: 9-1 answer 6 A. The final temperature could be higher than 0 without forming any ice. B. The final temperature could be higher than 0 with some water freezing. C. The final temperature could be 0 without forming any ice. D. The final temperature could be 0 with forming some ice. E. The final temperature could be 0 with all of water freezing. F. The final temperature could be below 0 without forming any ice. G. The final temperature could be below 0 with some water freezing. H. The final temperature could be below 0 with all of water freezing. Water cannot freeze above 0. If liquid water remains, it cannot be below 0.

7 Forming an assumption 7 The process that the aluminum could go through 1. Heat up to the final temperature The process that the water could go through 1. Cool down to 0, D to C 2. Freeze at 0, C to B 3. Cool down below 0, B to We can form assumptions by following these steps. Assumption I: aluminum heats up to 0, water cools down to 0 without forming any ice.

8 Quiz: 2 8 An aluminum cylinder of a mass m " = kg is removed from a liquid nitrogen bath, where it has been cooled to T ", = 196. The cylinder is immediately placed in an insulated cup containing water. The average specific heat of aluminum over this temperature range is c " = 653 J kg 6 K. Assumption I: The final temperature is T < = 0 without forming any ice. What is the heat the aluminum absorbs as it heated up to T < = 0 in Joules? If your answer is enter 1.234e6 or 1.234E6.

9 Quiz: 9-2 answer > J Q " = m " c " T " = m " c " T < T ", Q " = kg 653 J kg 6 K = > J

10 Quiz: 3 10 An aluminum cylinder is removed from a liquid nitrogen bath (below freezing point of water). The cylinder is immediately placed in an insulated cup containing m 1 = kg of water at T 1, = The specific heat of liquid water is c 1 = 4186 J kg 6 K. Assumption I: The final temperature is T < = 0 without forming any ice. What is the heat given off by the water as it cools down to T < = 0 without forming any ice in Joules? If your answer is enter 1.234e6 or 1.234E6.

11 Quiz: 9-3 answer C J Q 1 = m 1 c 1 T 1 = m 1 c 1 T < T 1, Q 1 = kg 4186 J kg 6 K = C J Q 1 is the heat added to the water. So the heat given off by water is Q 1.

12 Quiz: 4 12 An aluminum cylinder is removed from a liquid nitrogen bath (below freezing point of water). The cylinder is immediately placed in an insulated cup containing water. Under Assumption I, assuming that the final temperature is T < = 0 without forming any ice, we just calculated that the heat absorbed by aluminum is Q " = > J, and the heat given off by water is Q 1 = C J. What does this result alone imply? Choose all that apply. A. The final temperature could be higher than 0 without forming any ice. B. The final temperature could be 0 without forming any ice. C. The final temperature could be 0 with forming some ice. D. The final temperature could be 0 with all of water turning into ice. E. The final temperature could be below 0 with all of water turning into ice.

13 Quiz: 9-4 answer 13 The final temperature could be 0 with forming some ice. The final temperature could be 0 with all of water turning into ice. The final temperature could be below 0 with all of water turning into ice. When we assumed that the final temperature is T < = 0 without forming any ice, Q 1 < Q ", so the heat needed by the aluminum is greater than the heat that the water can supply to reach the assumed final condition. So, at least some water has to go through phase change to supply more heat to the aluminum.

14 Forming an assumption 14 The process that the aluminum could go through 1. Heat up to the final temperature The process that the water could go through 1. Cool down to 0, D to C 2. Freeze at 0, C to B 3. Cool down below 0, B to We can form assumptions by following these steps. Assumption II: aluminum heats up to 0, water cools down to 0 with all of water turning into ice.

15 Quiz: 5 15 An aluminum cylinder is removed from a liquid nitrogen bath (below freezing point of water). The cylinder is immediately placed in an insulated cup containing m 1 = kg of water at T 1, = The coefficient of latent heat of fusion for water is L 1, < = > J kg. Assumption II: the final temperature is T < = 0 with all of water turning into ice. What is the additional heat given off by the water in the process of all of it freezing at T < = 0 in Joules? If your answer is enter 1.234e6 or 1.234E6.

16 Quiz: 9-5 answer > J Q 1,< = m 1 L 1, < Q 1,< = kg > J kg = > J Q 1,< is the latent heat: heat added to melt ice, or heat given off by water as it freezes.

17 Quiz: 6 17 An aluminum cylinder is removed from a liquid nitrogen bath (below freezing point of water). The cylinder is immediately placed in an insulated cup containing water. Under assumption II, assuming that the final temperature is T < = 0 with all of water freezing, we just calculated that the heat absorbed by aluminum is Q " = > J, and the heat given off by water as it cools down is Q 1 = C J, and the heat given off by water as all of it freezes is Q 1,< = > J. What does this result alone imply? Choose all that apply. A. The final temperature could be 0 with some ice being formed. B. The final temperature could be 0 with all of water turning into ice. C. The final temperature could be below 0 with all of water turning into ice.

18 Quiz: 9-6 answer 18 The final temperature could be 0 with forming some ice. When we assumed that the final temperature is T < = 0 with all of water freezing, Q 1, < Q 1 > Q " C J > J = > J > > J So, the heat needed by the aluminum is smaller than the heat that the water can supply to reach the assumed final condition. Therefore not all water freezes. If the water is a liquid-ice mixture, the final temperature must be 0.

19 Quiz: 7 19 An aluminum cylinder is removed from a liquid nitrogen bath (below freezing point of water). The cylinder is immediately placed in an insulated cup containing water. The coefficient of latent heat of fusion for water is L 1, < = > J kg. Under assumption II, assuming that the final temperature is T < = 0 with all of water freezing, we just calculated that the heat absorbed by aluminum is Q " = > J, and the heat given off by water as it cools down is Q 1 = C J, and the heat given off by water as all of it freezes is Q 1,< = > J. Determine the amount of water that has actually frozen in kg.

20 Quiz: 9-7 answer kg Let m 1< the mass of ice formed Heat gained by aluminum = heat given off by water Q " = Q 1 + m 1< L 1, < m 1< = H IJH K L K, M = N.OP> NQR STU.QVC NQ W S CC.U NQ R S XY = kg