Good Morning Good Morning 04/20/2304/20/23

• Today we will be going over the answers Today we will be going over the answers to the questions on Worksheet 11.2a.to the questions on Worksheet 11.2a.

• Tomorrow we will be doing Lab 18 and Tomorrow we will be doing Lab 18 and Friday Lab 19.Friday Lab 19.

• So here we go.So here we go.

Worksheet 11.2a Worksheet 11.2a Specific Heat CapacitySpecific Heat Capacity

1. Complete the equation for 1. Complete the equation for finding energy transfer.finding energy transfer.

H = m H = m •• T T •• C C

2. What is the specific heat 2. What is the specific heat capacity for water?capacity for water?

• C = 4.184 C = 4.184 JJ

g g • • ºCºC

3. Calculate the energy 3. Calculate the energy required to raise the required to raise the

temperature of 750 g of water temperature of 750 g of water by 50.0 by 50.0 C.C.

H = H = • m = m = T = T = • C =C =

H = m H = m •• T T •• C C750g750g 4.184 4.184 JJ//g•ºCg•ºC50.0 50.0 CC

H = 156,900 J H = 156,900 J

= 156.9 kJ = 156.9 kJ

50.0 50.0 CC750g750g

4.184 4.184 JJ//g•ºCg•ºC

??

4. Calculate the specific heat of an 4. Calculate the specific heat of an unknown metal when 250 g of the metal unknown metal when 250 g of the metal

is cooled from 90.0is cooled from 90.0C to 28.0C to 28.0C when C when added to 133 g of water whose added to 133 g of water whose

temperature went form 22.0temperature went form 22.0C to 26.0C to 26.0C.C.

• This is just like the lab, where did the This is just like the lab, where did the heat go?heat go?

• Into the water so Into the water so HHmetalmetal = = HHww

4. Calculate the specific heat of an 4. Calculate the specific heat of an unknown metal when 250 g of the metal unknown metal when 250 g of the metal

is cooled from 90.0is cooled from 90.0C to 28.0C to 28.0C when C when added to 133 g of water whose added to 133 g of water whose

temperature went from 22.0temperature went from 22.0C to 26.0C to 26.0C.C.

H = ?H = ?• m = 250 gm = 250 gT = 62 ºCT = 62 ºC• C = ???C = ???

H = mH = mww •• T Tww •• C Cww133g133g 4.0 4.0 CC 4.184 4.184 JJ//g•ºCg•ºC =2225.9 J=2225.9 J

We need C all by itselfWe need C all by itself

H = m x C x T

m x Tm x T

H

m x T = C

2225.9 J = 0.1436 J/g· C

250.0 g 62C

5. Calculate the energy 5. Calculate the energy required to raise the required to raise the

temperature of 365.0 g of temperature of 365.0 g of water by 25.0 water by 25.0 C.C.

H =H =• m =m =T =T =• C =C =

H = m H = m •• T T •• C C365g365g 4.184 4.184 JJ//g•ºCg•ºC25.0 25.0 CC

H = 38,179 J H = 38,179 J = 38.2 kJ = 38.2 kJ

365g365g??

4.184 4.184 JJ//g•ºCg•ºC

25.0 25.0 CC

6. Calculate the specific heat (C) of an 6. Calculate the specific heat (C) of an unknown metal when 135.0 g of the unknown metal when 135.0 g of the

metal is cooled from 90.0 metal is cooled from 90.0 C to 25.0 C to 25.0 C by C by losing 3,396 Joules of energy.losing 3,396 Joules of energy.

H =H =• m =m =T =T =• C =C =

135 g135 g3,396 J3,396 J

????65.0 65.0 CC

H = m x C x H = m x C x TT

m x m x TTm x m x TT

HH

m x m x TT = C= C

3396 J3396 J = 0.387 J/g= 0.387 J/g· · CC

135.0 g135.0 g 6565CC

H = m x C x H = m x C x TT

7. Assume 126 Joules of heat is added 7. Assume 126 Joules of heat is added to 5.00 g of water originally at 23.0 to 5.00 g of water originally at 23.0 C. C. What would be the final temperature of What would be the final temperature of

the water?the water?

H =H =• m =m =T =T =• C =C =

5 g5 g126 J126 J

??m x Cm x Cm x Cm x C

HH

m x Cm x C = = ΔΔTT

126 J126 J = 6= 6 CC

5.0 g5.0 g

4.184 4.184 JJ//g•ºCg•ºC

4.184 4.184 JJ//g•ºCg•ºC

Final TempFinal Temp= 29= 29 CC

You don’t have to do 8, it isn’t a You don’t have to do 8, it isn’t a great question. Sorry Igreat question. Sorry I didn’t delete didn’t delete

it before handing it out.it before handing it out.

9. If 75.0 g of butane is burned, how 9. If 75.0 g of butane is burned, how much thermal energy, in kilojoules, is much thermal energy, in kilojoules, is

produced. produced.

Butane’s molar heat of combustion = Butane’s molar heat of combustion = 2859 kJ/mol2859 kJ/mol

75.0g 75.0g CC44HH1010 xx

1 mol C1 mol C44HH1010

xx58.14 g 58.14 g CC44HH1010

2859 kJ2859 kJ

1 mol 1 mol

CC44HH1010

= 3688 = 3688 kJkJ

10. How much thermal energy is produced 10. How much thermal energy is produced when one gallon of octane is burned? when one gallon of octane is burned?

Note: One gallon of octane has a mass of Note: One gallon of octane has a mass of

2,660 g and the molar heat of combustion is 2,660 g and the molar heat of combustion is 5450 kJ/mol5450 kJ/mol

2,660g 2,660g CC88HH1818 xx

1 mol C1 mol C88HH1818

xx114.26g 114.26g CC88HH1818

5450 kJ5450 kJ

1 mol 1 mol

CC88HH1818

= 126,900 kJ= 126,900 kJ

Lab 18 PRE-LAB EXERCISELab 18 PRE-LAB EXERCISE• In the space below, combine two of the equations In the space below, combine two of the equations

algebraically to obtain the third equation. Indicate the algebraically to obtain the third equation. Indicate the number of each reaction on the shorter lines.number of each reaction on the shorter lines.

• Hess’s Law says if you can combine 2 or more Hess’s Law says if you can combine 2 or more reactions to create a reaction you can add the reactions to create a reaction you can add the H’s for H’s for those reaction s to equal the those reaction s to equal the H of the final reaction.H of the final reaction.

• It is a little tricky but if you look close Reaction (2) is a It is a little tricky but if you look close Reaction (2) is a combination of Reactions (1) and (3)combination of Reactions (1) and (3)

(1) NaOH(1) NaOH(s) (s) Na Na++ + OH + OH–– H1H1

+ + + +(3) Na(3) Na++ + OH + OH–– + H + H++ + Cl + Cl–– H2O H2O(l) (l) + Na+ Na++ + Cl + Cl–– H3H3

== = =(2) NaOH(2) NaOH(s)(s) + H + H++ + Cl + Cl–– H2O H2O(l)(l) + Na + Na++ + Cl + Cl––

H2H2