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Homework: Page 495, problems 4-6 Page 495, questions 7-11.

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Presentation on theme: "Homework: Page 495, problems 4-6 Page 495, questions 7-11."— Presentation transcript:

1 Homework: Page 495, problems 4-6 Page 495, questions 7-11

2 4.If the temperature of 34.4 g of ethanol increases from 25.0 o C to 78.8 o C, how much heat has been absorbed by the ethanol?

3 q = c * m * T

4 4.If the temperature of 34.4 g of ethanol increases from 25.0 o C to 78.8 o C, how much heat has been absorbed by the ethanol? q = c * m * T = (2.44 J/g o )(34.4g)(44.4 o )

5 4.If the temperature of 34.4 g of ethanol increases from 25.0 o C to 78.8 o C, how much heat has been absorbed by the ethanol? q = c * m * T = (2.44 J/g o )(34.4g)(44.4 o ) = 4530

6 4.If the temperature of 34.4 g of ethanol increases from 25.0 o C to 78.8 o C, how much heat has been absorbed by the ethanol? q = c * m * T = (2.44 J/g o )(34.4g)(44.4 o ) = 4530 Units?

7 4.If the temperature of 34.4 g of ethanol increases from 25.0 o C to 78.8 o C, how much heat has been absorbed by the ethanol? q = c * m * T = (2.44 J/g o )(34.4g)(44.4 o ) = 4530 J

8 5.A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0 o C? c = J/g o

9 q = c * m * T

10 5.A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0 o C? c = J/g o q = c * m * T T = q / cm

11 5.A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0 o C? c = J/g o q = c * m * T T = q / cm = 276 J / (0.129 J/g o )( 4.50g)

12 5.A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0 o C? c = J/g o q = c * m * T T = q / cm = 276 J / (0.129 J/g o )( 4.50g) = 475 o C

13 5.A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0 o C? c = J/g o q = c * m * T T = q / cm = 276 J / (0.129 J/g o )( 4.50g) = 475 o C Final temp?

14 5.A 4.50g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0 o C? c = J/g o q = c * m * T T = q / cm = 276 J / (0.129 J/g o )( 4.50g) = 475 o C 25 o o = 500 o

15 6.155g of an unknown substance was heated from 25.0 o C to 40.0 o C. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.

16 q = c * m * T

17 6.155g of an unknown substance was heated from 25.0 o C to 40.0 o C. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table q = c * m * T c = q / m * T

18 6.155g of an unknown substance was heated from 25.0 o C to 40.0 o C. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table q = c * m * T c = q / m * T = 5696 J / (155g)(15 o C)

19 6.155g of an unknown substance was heated from 25.0 o C to 40.0 o C. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table q = c * m * T c = q / m * T = 5696 J / (155g)(15 o C) = 2.45J/g o C

20 6.155g of an unknown substance was heated from 25.0 o C to 40.0 o C. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table q = c * m * T c = q / m * T = 5696 J / (155g)(15 o C) = 2.45J/g o C So what is it?

21 6.155g of an unknown substance was heated from 25.0 o C to 40.0 o C. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table q = c * m * T c = q / m * T = 5696 J / (155g)(15 o C) = 2.45J/g o C ethanol

22 7.What is energy? The ability to do work or produce heat List two units for energy

23 7.What is energy? The ability to do work or produce heat List two units for energy

24 7.What is energy? The ability to do work or produce heat List two units for energy joules (J) calories (cal) Calories (Cal) kilocalories (kcal)

25 8.Kinetic or potential energy?

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32 9.What is the relationship between a calorie and a joule?

33 1 cal = j 1 j = cal

34 10.One lawn chair is aluminum, the other iron. Which is hotter?

35 Al c = j/g o C Fe c = j/g o C

36 10.One lawn chair is aluminum, the other iron. Which is hotter? Al c = j/g o C Fe c = j/g o C Iron is hotter.

37 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C?

38 1 cal = j

39 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? 1 cal = j 12 cal = 12(4.184) j

40 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? 1 cal = j 12 cal = 12(4.184) j 12 cal = j

41 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? q = c * m * T

42 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? q = c * m * T c = q / m * T

43 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? q = c * m * T c = q / m * T = j / (2.50g)(5 o C)

44 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? q = c * m * T c = q / m * T = j / (2.50g)(5 o C) = 4.02

45 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? q = c * m * T c = q / m * T = j / (2.50g)(5 o C) = 4.02 Units?

46 11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0 o C to 20 o C? q = c * m * T c = q / m * T = j / (2.50g)(5 o C) = 4.02 j/g o C


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