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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.

50 ˚C

B.

80 ˚C

C.

25 ˚C

D.

100 ˚C

E.

35 ˚C

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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.

50 ˚C

B.

80 ˚C

C.

25 ˚C

D.

100 ˚C

E.

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.

35 ˚C

B.

40 ˚C

C.

60 ˚C

D.

25 ˚C

E.

30 ˚C

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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.

35 ˚C

B.

40 ˚C

C.

60 ˚C

D.

25 ˚C

E.

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.

100 g

B.

25 g

C.

30 g

D.

50 g

E.

75 g

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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.

100 g

B.

25 g

C.

30 g

D.

50 g

E.

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.  

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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.

60 ˚C

B.

40 ˚C

C.

30 ˚C

D.

20 ˚C

E.

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.

60 ˚C

B.

40 ˚C

C.

30 ˚C

D.

20 ˚C

E.

50 ˚C

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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.

60 ˚C

B.

40 ˚C

C.

30 ˚C

D.

20 ˚C

E.

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.

70 ˚C

B.

40 ˚C

C.

30 ˚C

D.

60 ˚C

E.

50 ˚C

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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.

70 ˚C

B.

40 ˚C

C.

30 ˚C

D.

60 ˚C

E.

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.

70 ˚C

B.

40 ˚C

C.

30 ˚C

D.

60 ˚C

E.

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˚  

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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.

60 ˚C

B.

40 ˚C

C.

30 ˚C

D.

20 ˚C

E.

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.

60 ˚C

B.

40 ˚C

C.

30 ˚C

D.

20 ˚C

E.

50 ˚C

TWO  DIFFERENT  SUBSTANCES!    

Experimentally  the  final  temperature  is  determined  to  be  close  to  60˚.  

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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) 4.184 H2O(s) 2.03 Al(s) 0.89 C(s) 0.71 Fe(s) 0.45 Hg(l) 0.14 O2(g) 0.917

CH3CH2OH 2.46

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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

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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

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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

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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

SHmetal = 0.380 J g-1˚C-1

€

SH = −1608 J57.968 g• – 73.09 C

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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)

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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)

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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

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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)

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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