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**Page 495, problems 4-6 Page 495, questions 7-11**

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

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**4. If the temperature of 34. 4 g of ethanol increases from 25**

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?

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**If the temperature of 34. 4 g of ethanol increases from 25. 0oC to 78**

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

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**q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o)**

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)

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**q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530**

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

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**q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530 Units?**

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?

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**q = c * m * ∆T = (2.44 J/go)(34.4g)(44.4o) = 4530 J**

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

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**A 4. 50g nugget of pure gold absorbed 276 J of heat**

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 = J/go

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**A 4. 50g nugget of pure gold absorbed 276 J of heat**

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 = J/go q = c * m * ∆T

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**A 4. 50g nugget of pure gold absorbed 276 J of heat**

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 = J/go q = c * m * ∆T ∆T = q / cm

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**q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g)**

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 = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g)

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**q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC**

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 = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC

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**q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC**

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 = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC Final temp?

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**q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g)**

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 = J/go q = c * m * ∆T ∆T = q / cm = 276 J / (0.129 J/go)( 4.50g) = 475oC o + 475o = 500o

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**6. 155g of an unknown substance was heated from 25. 0oC to 40. 0oC**

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**

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

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**6. 155g of an unknown substance was heated from 25. 0oC to 40. 0oC**

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

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**q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC)**

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)

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**q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC) = 2.45J/goC**

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

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**q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC) = 2.45J/goC**

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?

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**q = c * m * ∆T c = q / m * ∆T = 5696 J / (155g)(15oC)**

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

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

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

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**joules (J) calories (cal) Calories (Cal) kilocalories (kcal)**

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

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**Kinetic or potential energy?**

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**Kinetic or potential energy?**

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**Kinetic or potential energy?**

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**Kinetic or potential energy?**

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**Kinetic or potential energy?**

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**Kinetic or potential energy?**

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**Kinetic or potential energy?**

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**What is the relationship between a calorie and a joule?**

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**What is the relationship between a calorie and a joule?**

1 cal = j 1 j = cal

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

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

Al c = j/goC Fe c = j/goC

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**Al c = 0.897 j/goC Fe c = 0.449 j/goC Iron is hotter.**

One lawn chair is aluminum, the other iron. Which is hotter? Al c = j/goC Fe c = j/goC Iron is hotter.

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**What is the specific heat of an unknown substance if a 2**

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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**What is the specific heat of an unknown substance if a 2**

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 = j

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**What is the specific heat of an unknown substance if a 2**

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 = j 12 cal = 12(4.184) j

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**What is the specific heat of an unknown substance if a 2**

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 = j 12 cal = 12(4.184) j 12 cal = j

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**What is the specific heat of an unknown substance if a 2**

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

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**What is the specific heat of an unknown substance if a 2**

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 c = q / m * ∆T

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**q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC)**

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 c = q / m * ∆T = j / (2.50g)(5oC)

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**q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02**

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 c = q / m * ∆T = j / (2.50g)(5oC) = 4.02

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**q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02 Units?**

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 c = q / m * ∆T = j / (2.50g)(5oC) = 4.02 Units?

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**q = c * m * ∆T c = q / m * ∆T = 50.208 j / (2.50g)(5oC) = 4.02 j/goC**

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 c = q / m * ∆T = j / (2.50g)(5oC) = 4.02 j/goC

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