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

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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

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?

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

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)

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

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?

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

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

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

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

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)

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

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?

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

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.

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

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

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)

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

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?

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

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

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

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)

Kinetic or potential energy?

Kinetic or potential energy?

Kinetic or potential energy?

Kinetic or potential energy?

Kinetic or potential energy?

Kinetic or potential energy?

Kinetic or potential energy?

What is the relationship between a calorie and a joule?

What is the relationship between a calorie and a joule?
1 cal = j 1 j = cal

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

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

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.

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?

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

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

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

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

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

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)

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

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?

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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