9/15/12 Introduction to Thermodynamics
So what kind of intuition do we have about heat and temperature
and energy?
Discuss the DCI.
Introduction to Thermodynamics
DCI15.1.
Two containers of water are at 20
˚C initially. One contains 50 mLs and the other
100 mLs. They are each heated with the same
source of heat for the same amount of time. If
the final temperature of the 50 mLs sample is 50
˚C what would be the final temperature of the
100 mLs sample?
A.
B.
C.
D.
E.
50 ˚C
80 ˚C
25 ˚C
100 ˚C
35 ˚C
1 9/15/12 Introduction to Thermodynamics
DCI15.1. Two containers of water are at 20 ˚C
initially. One contains 50 mLs and the other 100
mLs. They are each heated with the same source
of heat for the same amount of time. If the final
temperature of the 50 mLs sample is 50 ˚C what
would be the final temperature of the 100 mLs
sample?
A.
B.
C.
D.
E.
50 ˚C
80 ˚C
25 ˚C
100 ˚C
35 ˚C
Same amount of heat to both beakers, but different mass. ∆T = 30˚ for beaker on the leA, so ∆T is half or 15˚. Mass α 1/∆T Introduction to Thermodynamics
Two containers each have 50 mLs of water at 20 ˚C initially. They are each heated with the same
source of heat. One is heated for ten minutes
and the other for five minutes. If the container
that was heated for five minutes has a final
temperature 30 ˚C what would be the final
temperature of the other sample?
A.
B.
C.
D.
E.
35 ˚C
40 ˚C
60 ˚C
25 ˚C
30 ˚C
2 9/15/12 Introduction to Thermodynamics
Two containers each have 50 mLs of water at 20 ˚C initially. They are each heated with the same
source of heat. One is heated for ten minutes
and the other for five minutes. If the container
that was heated for five minutes has a final
temperature 30 ˚C what would be the final
temperature of the other sample?
A.
B.
C.
D.
E.
35 ˚C
40 ˚C
60 ˚C
25 ˚C
30 ˚C
Both beakers contain the same amount of water. Twice the heat to one. ∆T is 10˚ for smaller amount of heat, than ∆T = 20˚ for larger amount. Q(heat) α ∆T Introduction to Thermodynamics
Two containers of water are at 20 ˚C initially.
One contains 50 g of water and is heated by a
source for a specified time to a final temperature
of 30 ˚C. The second container has an unknown
amount of water and is heated with the same
source to 30 ˚C. However, it takes twice as long
to get to this final temperature. How much water
is in this container?
A.
B.
C.
D.
E.
100 g
25 g
30 g
50 g
75 g
3 9/15/12 Introduction to Thermodynamics
Two containers of water are at 20 ˚C initially.
One contains 50 g of water and is heated by a
source for a specified time to a final temperature
of 30 ˚C. The second container has an unknown
amount of water and is heated with the same
source to 30 ˚C. However, it takes twice as long
to get to this final temperature. How much water
is in this container?
A.
B.
C.
D.
E.
100 g
25 g
30 g
50 g
75 g
Twice the heat is added to one beaker to reach the same final temperature (∆T). So the beaker must have twice the mass. Q(heat) α Mass Introduction to Thermodynamics
So we have established the following relaNonships; Mass α 1/∆T q(heat) α ∆T q(heat) α mass So q(heat) α mass ·∙ ∆T Heat is directly proporNonal to the mass Nmes the change in temperature. 4 9/15/12 Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 50 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 50 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
5 9/15/12 Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 50 mLs of
water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
qhot water + qcold water = 0
qhot water = –qcold water masshot water ·∙ ∆Thot water = –masscold water ·∙ ∆Tcold water 50. g ·∙ ∆Thot water = –50. g ·∙ ∆Tcold water 50. g ·∙ (Tfinal – 80.0˚) = –50. g ·∙ (Tfinal – 20.0˚) 2Tfinal = 100˚ Tfinal = 50˚ Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 100 mLs of water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
70 ˚C
40 ˚C
30 ˚C
60 ˚C
50 ˚C
6 9/15/12 Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 100 mLs of water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
70 ˚C
40 ˚C
30 ˚C
60 ˚C
50 ˚C
Introduction to Thermodynamics
50 mLs of water at 80 ˚C is added to 100 mLs of water at 20 ˚C. What would be the final
temperature?
A.
B.
C.
D.
E.
70 ˚C
40 ˚C
30 ˚C
60 ˚C
50 ˚C
qhot water = –qcold water masshot water ·∙ ∆Thot water = –masscold water ·∙ ∆Tcold water 50. g ·∙ ∆Thot water = –100. g ·∙ ∆Tcold water 50. g ·∙ (Tfinal – 80.0˚) = –100. g ·∙ (Tfinal – 20.0˚) (Tfinal – 80.0˚) = –2 ·∙ (Tfinal – 20.0˚) 3Tfinal = 120˚ Tfinal = 40˚ 7 9/15/12 Introduction to Thermodynamics
50 g of water at 80 ˚C is added to 50 g of ethyl
alcohol at 20 ˚C. What would be the
approximate final temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
Introduction to Thermodynamics
50 g of water at 80 ˚C is added to 50 g of ethyl
alcohol at 20 ˚C. What would be the
approximate final temperature?
A.
B.
C.
D.
E.
60 ˚C
40 ˚C
30 ˚C
20 ˚C
50 ˚C
TWO DIFFERENT SUBSTANCES! Experimentally the final temperature is determined to be close to 60˚. 8 9/15/12 Introduction to Thermodynamics
q(heat) α mass · ∆T
How do we make this an equality?
We must introduce a constant….in this case the
constant is called the specific heat, SH,
q(heat) = mass · SH · ∆T
Specific heat is the amount of heat required to raise the
temperature of 1 gram of a substance 1 ˚C.
Introduction to Thermodynamics
Specific Heats of Substances
Compound
Specific Heat
(J ˚C-1g-1)
H2O(l)
H2O(s)
Al(s)
C(s)
Fe(s)
Hg(l)
O2(g)
CH3CH2OH
4.184
2.03
0.89
0.71
0.45
0.14
0.917
2.46
9 9/15/12 Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
A 175 g sample of water, initially at 23.45 ˚C absorbs some
heat. The final temperature of the sample after absorbing the
heat is 26.85 ˚C. Calculate the amount of heat absorbed by the
sample of water. (NOTE: The specific heat for water is 4.184 J
g-1 ˚C-1.)
q = masswater · SHwater · ∆Twater
Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
A 175 g sample of water, initially at 23.45 ˚C absorbs some
heat. The final temperature of the sample after absorbing the
heat is 26.85 ˚C. Calculate the amount of heat absorbed by the
sample of water. (NOTE: The specific heat for water is 4.184 J
g-1 ˚C-1.)
q = masswater · SHwater · ∆Twater
q = 175. g · 4.184 J g-1˚C-1 · (26.85˚ – 23.45˚)
q = 2.49 x 103 J
10 9/15/12 Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
A piece of iron weighing 80.0 g initially at a temperature of
92.6 ˚C released the same amount of heat to the 175 g sample
of water in DCI16.4. Assume the final temperature of the
metal is the same as the final temperature of the water in
DCI16.4. What is the specific heat for iron?
Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
A piece of iron weighing 80.0 g initially at a temperature of
92.6 ˚C released the same amount of heat to the 175 g sample
of water in DCI16.4. Assume the final temperature of the
metal is the same as the final temperature of the water in
DCI16.4. What is the specific heat for iron?
The metal is absorbing 2.49 x 103 J of heat
11 9/15/12 Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
A piece of iron weighing 80.0 g initially at a temperature of
92.6 ˚C released the same amount of heat to the 175 g sample
of water in DCI16.4. Assume the final temperature of the
metal is the same as the final temperature of the water in
DCI16.4. What is the specific heat for iron?
The metal is absorbing 2.49 x 103 J of heat
q = massFe · SHFe · ∆TFe
Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
A piece of iron weighing 80.0 g initially at a temperature of 92.6 ˚C
released the same amount of heat to the 175 g sample of water in
DCI16.4. Assume the final temperature of the metal is the same as the
final temperature of the water in DCI16.4. What is the specific heat for
iron?
The metal is absorbing 2.49 x 103 J of heat
q = massFe · SHFe · ∆Tfe
–2.49 x 103 J = 80.0 g · SHFe · (26.85 ˚C - 92.6 ˚C) SHFe = –2.49 x 103 J /(80.0 g · (-65.75 ˚C)) SHFe = 0.473 J g-1 ˚C-1
12 9/15/12 Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
The four pictures shown below summarize an experiment. A zinc
cylinder of mass 57.968 g was placed in boiling water at 100 ˚C then
plunged into a beaker containing 169.340 g of water at 24.64 ˚C. The
temperature of the water and zinc cylinder finally levels off at 26.91 ˚C.
Calculate the specific heat of zinc metal.
Check out the movie
Check out the movie
Introduction to Thermodynamics
q(heat) = mass · SH · ∆T
The four pictures shown below summarize an experiment. A zinc cylinder of mass
57.968 g was placed in boiling water at 100 ˚C then plunged into a beaker containing
169.340 g of water at 24.64 ˚C. The temperature of the water and zinc cylinder finally
levels off at 26.91 ˚C. Calculate the specific heat of zinc metal.
qmetal = –qwater
qwater = masswater · SHwater · ∆Twater
qwater = 169.340 g · 4.814 J g-1˚C-1 · (26.91 – 24.64˚)
qwater = 1608. J
qmetal = –1608. J
massmetal · SHmetal · ∆Tmetal = –1608. J
57.968. g · SHmetal · (26.91 – 100.0˚) = –1608. J
SH =
−1608 J
57.968 g• – 73.09 C
SHmetal = 0.380 J g-1˚C-1
€
13 9/15/12 Introduction to Thermodynamics
IMPORTANT RELATIONSHIPS
q = mass*SH*∆T (can use qhot, qcold, qmetal, qrxn)
qhot = -qcold
qmetal = -qcold
Coffee Cup Calorimeter
•  q (lost by reaction) = -q (gained by water)
•  q (lost by reaction) = -q (H2O) = m SH (H2O) ∆t (H2O)
14 9/15/12 Introduction to Thermodynamics
A 100.00 ml sample of 0.200 M CsOH is mixed
with 100. mL of 0.200 M HCl in an OSU
calorimeter the following reaction occurs
CsOH(aq) + HCl(aq) → CsCl(aq) + H2O(l)
The temperature before mixing of both solutions
is 24.30 ˚C. After mixing the final temperature is
25.6 ˚C. The heat capacity of the calorimeter is
50. J ˚C-1 and the specific heat of the solution is
4.20 J g-1˚C-1. Calculate the heat released in the
reaction.
Introduction to Thermodynamics
A 100.00 ml sample of 0.200 M CsOH is mixed with
100. mL of 0.200 M HCl in an OSU calorimeter the
following reaction occurs
CsOH(aq) + HCl(aq) → CsCl(aq) + H2O(l)
The temperature before mixing of both solutions is
24.30 ˚C. After mixing the final temperature is 25.6 ˚C.
The heat capacity of the calorimeter is 50. J ˚C-1 and
the specific heat of the solution is 4.184 J g-1˚C-1.
Calculate the heat released in the reaction. (Assume
the density of the solution is 1.00 g mL-1)
qrxn = -(qsolution + qcalorimeter)
15 9/15/12 Introduction to Thermodynamics
A 100.00 ml sample of 0.200 M CsOH is mixed with
100. mL of 0.200 M HCl in an OSU calorimeter the
following reaction occurs
CsOH(aq) + HCl(aq) → CsCl(aq) + H2O(l)
The temperature before mixing of both solutions is
24.30 ˚C. After mixing the final temperature is 25.6 ˚C.
The heat capacity of the calorimeter is 50. J ˚C-1 and
the specific heat of the solution is 4.184 J g-1˚C-1.
Calculate the heat released in the reaction. (Assume
the density of the solution is 1.00 g mL-1)
qrxn = -(qsolution + qcalorimeter)
qrxn = -(masssoln*SHsoln*∆Tsolution + HCcal*∆Tcalorimeter)
Introduction to Thermodynamics
A 100.00 ml sample of 0.200 M CsOH is mixed with 100. mL of
0.200 M HCl in an OSU calorimeter the following reaction occurs
CsOH(aq) + HCl(aq) → CsCl(aq) + H2O(l)
The temperature before mixing of both solutions is 24.30 ˚C.
After mixing the final temperature is 25.6 ˚C. The heat capacity
of the calorimeter is 50. J ˚C-1 and the specific heat of the
solution is 4.184 J g-1˚C-1. Calculate the heat released in the
reaction. (Assume the density of the solution is 1.00 g
mL-1)
qrxn = -(qsolution + qcalorimeter)
qrxn = -(200.g* 4.184 J g-1˚C-1 *(25.6 – 24.30) + 50. J ˚C-1 * (25.6
– 24.30) )
qrxn = -(200.g* 4.184 J g-1˚C-1 *(1.3 ˚C) + 50. J ˚C-1 * (1.3 ˚C) )
qrxn = -1153 J qrxn = -1153 J/0.02 mol = -57.6 kJ/mol
16 9/15/12 Introduction to Thermodynamics
A 0.692 g sample of glucose, C6H12O6, is burned in a
constant volume bomb calorimeter and the following
reaction occurs
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
The temperature change of the water and the
calorimeter is 1.80 ˚C. The calorimeter contains 1.05 kg
of water and the dry calorimeter has a heat capacity of
650 J ˚C-1 and the specific heat of the water is 4.184 J
g-1˚C-1. Calculate the heat released in the reaction.
Introduction to Thermodynamics
A 0.692 g sample of glucose, C6H12O6, is burned in a
constant volume bomb calorimeter and the following
reaction occurs
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
The temperature change of the water and the
calorimeter is 1.80 ˚C. The calorimeter contains 1.05 kg
of water and the dry calorimeter has a heat capacity of
650 J ˚C-1 and the specific heat of the water is 4.184 J
g-1˚C-1. Calculate the heat released in the reaction.
qrxn = -(qwater + qcalorimeter)
17 9/15/12 Introduction to Thermodynamics
A 0.692 g sample of glucose, C6H12O6, is burned in a constant
volume bomb calorimeter and the following reaction occurs
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
The temperature change of the water and the calorimeter is 1.80
˚C. The calorimeter contains 1.05 kg of water and the dry
calorimeter has a heat capacity of 650 J ˚C-1 and the specific heat
of the water is 4.184 J g-1˚C-1. Calculate the heat released in the
reaction.
qrxn = -(qwater + qcalorimeter)
qrxn = -(1050.g* 4.184 J g-1˚C-1 *(1.8 ˚C) + 650. J ˚C-1 * (1.8 ˚C) )
qrxn = –9078 J
qrxn = -9078 J/0.692 g C6H12O6 *(180 g C6H12O6 /1 mol C6H12O6 ) = -2361 kJ/
mol
Introduction to Thermodynamics
IMPORTANT RELATIONSHIPS
q = mass*SH*∆T (can use qhot, qcold, qmetal, qrxn)
qhot = -qcold
qhot = -(qcold + qcalorimeter) OSU calorimeter
qmetal = -qcold
qmetal = -(qcold + qcalorimeter) OSU calorimeter
qrxn = -(qsolution + qcalorimeter) OSU calorimeter
qrxn = -(qwater + qcalorimeter) bomb calorimeter
18