1. Use the guided notes doc calorimetry and latent heat. Follow the PowerPoint lecture to answer the questions on the guided questions. The guided questions were collected at the end of the classroom for points.

2. Warm UP Q = mcΔT 1)A.050 Kg metal bolt is heated to an unknown initial temperature. It is then dropped into a calorimeter containing 0.15 kg of water with initial temperature of 21.0 °C. The bolt and the water then reach a final temperature of 25 °C. Find the initial temperature of the metal bolt. Metal has a specific heat capacity of 900 J/kg°C Water has a specific heat capacity of 4,200 J/kg°C

3. Given metal : m =.05 Kg T f = 25°C c = 900 J/kg°C Unknown: T i metal = ??? Given water : m =.15 kg T i =21°C T f =25°C c = 4,200 J/kg°C

4. |-Q metal | =| Q h20 | mcΔT =mcΔT (.05kg)(900 J/kg°C) ( T im - 25 °C) = (.15kg)(4,200 J/g°C)( 25°C -21°C) (.05kg)(900 J/kg°C) ( T i m - 25 °C) = 2,520 J T i m - 25 °C = 2,520 J /(.05kg)(900 J/kg°C) T i m = [2,520 J / (.05kg)(900 J/kg°C)] + 25 °C T i m = 81 °C

5. Monday Review Monday we’re meeting in the amphitheater May 9 th Unit test Test = Wednesday

6. 2) What will require more heat in order to raise the temperature by 1°C? A) 1 gram of Water Vapor ( steam) B) 1 gram of solid water ( ice) C) 1 gram of liquid water Explain why did you choose this answer ?

7. Warm UP What will require more heat in order to raise the temperature by 1°C? A) 1 gram of Water Vapor ( steam) B) 1 gram of solid water ( ice) C) 1 gram of liquid water Why did you choose this answer ? Water has the highest Specific heat capacity of 4.18 J thus requires more heat to raise 1 gram by 1°C.

8. Phase changes and latent Heat The substance will go through a phase change. The three phases of matter are Solid, Liquid and Vapor.

10. In these phase changes, heat transfer does not cause a temperature change, Q = mcΔT doesn't apply to phase changes. Relevant equation

11. L stands for the latent heat of transformation

12. Heating of materials Q = mcΔT c  specific heat capacity of the material Phase change of Materials Q =m L f Q =m L v L f - latent heat of fusion ( Solid  liquid) L v - latent heat of vaporization (liquid  gas)

13. Latent heat Unit  J/Kg L f and L v can be found from the chart

14. no The plateaus are also called phase changes That is when there is no change in temperature.

16. Melting Q =m L f Boiling Q =m L v

17. 3) Label which formula do you use for : a)Freezing, Q = _________ b) Melting, Q=_________ c) Boiling, Q=________ d) Condensing, Q =_______ e) Heating up, Q=______ f) Cooling down, Q=______ g) Vaporization, Q=________ h) Fusion, Q=___________ Q =m L f Q= m L v Q=m c  T

18. Q =m L f Q= m L v Q=m c  T Labels which formula do you use for : Q= m L f 1) Freezing Q= m L f Q= m L f 2) Melting Q= m L f Q= m L v 3) Boiling Q= m L v Q= m L v 4) Condensing Q= m L v Q=m c  T 5) Heating up Q=m c  T Q=m c  T 6) Cooling down Q=m c  T Q= m L v 7) Vaporization Q= m L v Q =m L f 8) Fusion Q =m L f

19. 4) How much heats required to convert 10 g of ice at - 10 °C into steam at a temperature of 110 °C ? c ice = 2.1 J/g°C, c steam = 2.2 J/g°C, c h20 = 4.18 J/g°C, L f = 334 J/g, L v = 225 J/g)

20. Q total Total energy required to change solid ice at -10°C into 110°C steam Q -ice =(10)(2.1 J/g°C)(10°C)=210 J Q fusion =(10g)(334 j/g) = 3,340 J Q liquid =(10g) ( 4.18 J/g°C)(100°C) = 4,180 J Q vaporization = (10g) (225 J/g) = 2,250 J Q steam =(10g) (2.2 j/g°c) (10°C)=220 J Q total =210 J+3,340 J+4,180 J+2,250 J+220 J = 10,200 J Q total =10,200 J

21. 50 grams 100 °C steam cools to temperature of the skin of 37 °C. The water is now in equilibrium with the skin. What is the final temperature? 37 °C What is the initial temperature of steam? 100 °C Is there a phase change? yes – from steam to liquid water (Condensation) How much total energy was given off for steam to reach the temperature of the skin? What formulas will you use ? Q Lv = mLv water  L v = 225 J/g Q =mc (T final - T initial ) Q total = Q 1 +Q 2 …

22. From 100 °C steam to 100 °C liquid (change of phase) Q = mL v (50g)(225 J/g°C) = 11,250 J From 100 °C liquid water to 37°C Q = m c  T(50g) ( 4.18 J/g °C) (63 °C ) = 13,167 J Q total = Q 1 +Q 2 11,250 J + 13,167 J = 24,417 J The 50 grams of steam liberates 24,417 J to reach the temperature of the skin ( 37 °C)

23. Both water at 100 °C and steam at 100 °C can cause serious burns. Given the same mass, Is a burn produced by steam likely to be: A) more severe B) less severe C) the same Support your answer !

24. A) A) The steam is more severe 100 °C water cools to final temperature of the skin (37 °C or 98.7 °F) 100 °C Water transfer its thermal energy (heat, Q) to the skin 100 °C steam 1 st steam gives away thermal energy ( heat, Q) for phase change ; condensation to water at 100 °C 2 nd condensed water at 100 °C cools to body temp Thus Steam transfers more energy to the skin.

25. What happens to the temperature when heat is added to convert a substance from one phase to another ??? Latent, L is the thermal energy ( heat; Q ) required to change ( 1kg or 1 gram ) of a substance from one phase to another. During the conversion from one phase to another, the temperature of the system remains the same(constant) Heat to change phase = mass x latent heat Q = mL