Transcript Wednesday, September 5

Homework:
Page 495, problems 4-6
Page 495, questions 7-11
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?
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
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
o
o
= (2.44 J/g )(34.4g)(44.4 )
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
o
o
= (2.44 J/g )(34.4g)(44.4 )
= 4530
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
o
o
= (2.44 J/g )(34.4g)(44.4 )
= 4530
Units?
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
o
o
= (2.44 J/g )(34.4g)(44.4 )
= 4530 J
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
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
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
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
o
= 276 J / (0.129 J/g )( 4.50g)
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
o
= 276 J / (0.129 J/g )( 4.50g)
o
= 475 C
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
o
= 276 J / (0.129 J/g )( 4.50g)
o
= 475 C
Final temp?
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
o
= 276 J / (0.129 J/g )( 4.50g)
o
o
o
o
= 475 C 25 + 475 = 500
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. 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. 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
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
o
= 5696 J / (155g)(15 C)
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
o
= 5696 J / (155g)(15 C)
o
= 2.45J/g C
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
o
= 5696 J / (155g)(15 C)
o
= 2.45J/g C So what is it?
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
o
= 5696 J / (155g)(15 C)
o
= 2.45J/g C
ethanol
7. What is energy?
The ability to do work or
produce heat
List two units for energy
7. What is energy?
The ability to do work or
produce heat
List two units for energy
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)
8. Kinetic or potential energy?
8. Kinetic or potential energy?
8. Kinetic or potential energy?
8. Kinetic or potential energy?
8. Kinetic or potential energy?
8. Kinetic or potential energy?
8. Kinetic or potential energy?
9. What is the relationship
between a calorie and a joule?
9. What is the relationship
between a calorie and a joule?
1 cal = 4.184 j
1 j = 0.239 cal
10.One lawn chair is aluminum, the
other iron. Which is hotter?
10.One lawn chair is aluminum, the
other iron. Which is hotter?
o
j/g C
Al c = 0.897
o
Fe c = 0.449 j/g C
10.One lawn chair is aluminum, the
other iron. Which is hotter?
o
j/g C
Al c = 0.897
o
Fe c = 0.449 j/g C
Iron is hotter.
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?
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
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
12 cal = 12(4.184) j
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
12 cal = 12(4.184) j
12 cal = 50.208 j
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
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
c = q / m * ∆T
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
c = q / m * ∆T
= 50.208 j / (2.50g)(5oC)
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
c = q / m * ∆T
= 50.208 j / (2.50g)(5oC)
= 4.02
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
c = q / m * ∆T
= 50.208 j / (2.50g)(5oC)
= 4.02
Units?
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
c = q / m * ∆T
= 50.208 j / (2.50g)(5oC)
= 4.02 j/goC