homework: page 495, problems 4-6 page 495, questions 7-11
TRANSCRIPT
![Page 1: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/1.jpg)
Homework:
Page 495, problems 4-6
Page 495, questions 7-11
![Page 2: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/2.jpg)
4. If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol?
![Page 3: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/3.jpg)
4. If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol?
q = c * m * ∆T
![Page 4: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/4.jpg)
4. If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol?
q = c * m * ∆T
= (2.44 J/go)(34.4g)(44.4o)
![Page 5: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/5.jpg)
4. If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol?
q = c * m * ∆T
= (2.44 J/go)(34.4g)(44.4o)
= 4530
![Page 6: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/6.jpg)
4. If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol?
q = c * m * ∆T
= (2.44 J/go)(34.4g)(44.4o)
= 4530 Units?
![Page 7: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/7.jpg)
4. If the temperature of 34.4 g of ethanol increases from 25.0oC to 78.8oC, how much heat has been absorbed by the ethanol?
q = c * m * ∆T
= (2.44 J/go)(34.4g)(44.4o)
= 4530 J
![Page 8: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/8.jpg)
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 25.0oC? c = 0.129 J/go
![Page 9: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/9.jpg)
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 25.0oC? c = 0.129 J/go
q = c * m * ∆T
![Page 10: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/10.jpg)
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 25.0oC? c = 0.129 J/go
q = c * m * ∆T
∆T = q / cm
![Page 11: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/11.jpg)
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 25.0oC? c = 0.129 J/go
q = c * m * ∆T
∆T = q / cm
= 276 J / (0.129 J/go)( 4.50g)
![Page 12: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/12.jpg)
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 25.0oC? c = 0.129 J/go
q = c * m * ∆T
∆T = q / cm
= 276 J / (0.129 J/go)( 4.50g)
= 475oC
![Page 13: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/13.jpg)
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 25.0oC? c = 0.129 J/go
q = c * m * ∆T
∆T = q / cm
= 276 J / (0.129 J/go)( 4.50g)
= 475oC Final temp?
![Page 14: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/14.jpg)
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 25.0oC? c = 0.129 J/go
q = c * m * ∆T
∆T = q / cm
= 276 J / (0.129 J/go)( 4.50g)
= 475oC 25o + 475o = 500o
![Page 15: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/15.jpg)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
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6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
q = c * m * ∆T
![Page 17: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/17.jpg)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
q = c * m * ∆T
c = q / m * ∆T
![Page 18: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/18.jpg)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
q = c * m * ∆T
c = q / m * ∆T
= 5696 J / (155g)(15oC)
![Page 19: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/19.jpg)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
q = c * m * ∆T
c = q / m * ∆T
= 5696 J / (155g)(15oC)
= 2.45J/goC
![Page 20: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/20.jpg)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
q = c * m * ∆T
c = q / m * ∆T
= 5696 J / (155g)(15oC)
= 2.45J/goC So what is it?
![Page 21: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/21.jpg)
6. 155g of an unknown substance was heated from 25.0oC to 40.0oC. It absorbed 5696J of energy. What is its specific heat? Identify it from those listed in table 16-2.
q = c * m * ∆T
c = q / m * ∆T
= 5696 J / (155g)(15oC)
= 2.45J/goC ethanol
![Page 22: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/22.jpg)
7. What is energy?
The ability to do work or produce heatList two units for energy
![Page 23: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/23.jpg)
7. What is energy?
The ability to do work or produce heatList two units for energy
![Page 24: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/24.jpg)
7. What is energy?
The ability to do work or produce heatList two units for energy
joules (J)calories (cal)Calories (Cal)kilocalories (kcal)
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8. Kinetic or potential energy?
![Page 26: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/26.jpg)
8. Kinetic or potential energy?
![Page 27: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/27.jpg)
8. Kinetic or potential energy?
![Page 28: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/28.jpg)
8. Kinetic or potential energy?
![Page 29: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/29.jpg)
8. Kinetic or potential energy?
![Page 30: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/30.jpg)
8. Kinetic or potential energy?
![Page 31: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/31.jpg)
8. Kinetic or potential energy?
![Page 32: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/32.jpg)
9. What is the relationship between a calorie and a joule?
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9. What is the relationship between a calorie and a joule?
1 cal = 4.184 j1 j = 0.239 cal
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10.One lawn chair is aluminum, the other iron. Which is hotter?
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10.One lawn chair is aluminum, the other iron. Which is hotter?
Al c = 0.897 j/goCFe c = 0.449 j/goC
![Page 36: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/36.jpg)
10.One lawn chair is aluminum, the other iron. Which is hotter?
Al c = 0.897 j/goCFe c = 0.449 j/goC
Iron is hotter.
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11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
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11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
1 cal = 4.184 j
![Page 39: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/39.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
1 cal = 4.184 j12 cal = 12(4.184) j
![Page 40: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/40.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
1 cal = 4.184 j12 cal = 12(4.184) j
12 cal = 50.208 j
![Page 41: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/41.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
q = c * m * ∆T
![Page 42: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/42.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
q = c * m * ∆Tc = q / m * ∆T
![Page 43: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/43.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
q = c * m * ∆Tc = q / m * ∆T = 50.208 j / (2.50g)(5oC)
![Page 44: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/44.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
q = c * m * ∆Tc = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02
![Page 45: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/45.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
q = c * m * ∆Tc = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02 Units?
![Page 46: Homework: Page 495, problems 4-6 Page 495, questions 7-11](https://reader036.vdocuments.mx/reader036/viewer/2022081513/551a2e08550346cb358b4f31/html5/thumbnails/46.jpg)
11.What is the specific heat of an unknown substance if a 2.50g sample releases 12.0 cal as its temperature changes from 25.0oC to 20oC?
q = c * m * ∆Tc = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02 j/goC