A.
Specific Heat of a Metal
Trial 1 Trial 2
1. Mass of metal (𝑔) 10 𝑔
2. Temperature of metal
98°𝐶
(boiling water) (°𝐶)
3. Mass of calorimeter (𝑔) 31.22 𝑔
4. Mass of calorimeter +
49.65 𝑔
water (𝑔)
5. Mass of water (𝑔) 18.43 𝑔
6. Temperature of water in
32°𝐶
calorimeter (°𝐶)
7. Maximum temperature of
metal and water from 36°𝐶
graph (°𝐶)
Calculations for Specific Heat
of a Metal
1. Temperature change of
4°𝐶
water, ∆𝑇 (°𝐶)
2. Heat gained by water (J) 308.15 J
3. Temperature change of
-62°𝐶
metal, ∆𝑇 (°𝐶)
4. Specific heat of metal
0.4970 𝐽⁄𝑔 ∙ °𝐶
(𝐽⁄𝑔 ∙ °𝐶 )
5. Average specific heat of
metal (𝐽⁄𝑔 ∙ °𝐶 )
B. Enthalpy (Heat) of Neutralization for an Acid-Base Reaction
𝐻𝐶𝑙 + 𝑁𝑎𝑂𝐻 𝐻𝑁𝑂3 + 𝑁𝑎𝑂𝐻
Trial 1 Trial 2 Trial 1 Trial 2
1. Volume of Acid (𝑚𝐿) 50 𝑚𝐿 50 𝑚𝐿
2. Temperature of Acid
30°𝐶 32°𝐶
(°𝐶)
3. Volume of 𝑁𝑎𝑂𝐻 (𝑚𝐿) 50 𝑚𝐿 50 𝑚𝐿
4. Temperature of 𝑁𝑎𝑂𝐻
22°𝐶 22°𝐶
(°𝐶)
5. Exact molar
concentration of 𝑁𝑎𝑂𝐻 0.9832 𝑚𝑜𝑙/𝐿 0.9832 𝑚𝑜𝑙/𝐿
(𝑚𝑜𝑙/𝐿)
6. Maximum temperature
38°𝐶 32°𝐶
from graph (°𝐶)
Calculations for Enthalpy (Heat) of Neutralization for an Acid-Base Reaction
1. Average initial
temperature of acid and 26°𝐶 27°𝐶
𝑁𝑎𝑂𝐻 (°𝐶)
2. Temperature change, ∆𝑇
12°𝐶 5°𝐶
(°𝐶)
3. Volume of final mixture
100 𝑚𝐿 100 𝑚𝐿
(𝑚𝐿)
4. Mass of final mixture (𝑔)
(Assume the density of the 100 𝑔 100 𝑔
solution is 1.0 𝑔/𝑚𝐿.)
5. Specific heat of
4.18 𝐽⁄𝑔 ∙ °𝐶 4.18 𝐽⁄𝑔 ∙ °𝐶
mixture
6. Heat evolved (𝐽) 5016 𝐽 2090 𝐽
7. Moles of 𝑂𝐻 − reacted,
the limiting reactant 0.04916 𝑚𝑜𝑙 0.04916 𝑚𝑜𝑙
(𝑚𝑜𝑙)
8. Moles of 𝐻2 𝑂 formed
0.04916 𝑚𝑜𝑙 0.04916 𝑚𝑜𝑙
(𝑚𝑜𝑙)
9. ∆𝐻𝑛 (𝑘𝐽⁄𝑚𝑜𝑙 𝐻2 𝑂) 𝑘𝐽 𝑘𝐽
−102.0341 −42.5142
𝑚𝑜𝑙𝐻2 𝑂 𝑚𝑜𝑙𝐻2 𝑂
10. Average
∆𝐻𝑛 (𝑘𝐽⁄𝑚𝑜𝑙 𝐻2 𝑂)
C. Enthalpy (Heat) of Solution for the Dissolution of a Salt
Trial 1 Trial 2
1. Mass of salt (𝑔) 5g
2. Moles of salt (mol) 0.0806 mol
3. Mass of calorimeter (𝑔) 31.22 g
4. Mass of calorimeter +
49.65 g
water (𝑔)
5. Mass of water (𝑔) 18.43 g
6. Initial Temperature of
30 °C
water (°𝐶)
7. Final Temperature of
20 °C
mixture from graph (°C)
Calculations for Enthalpy (Heat) of Solution for the Dissolution of a Salt
1. Temperature change of
-10°𝐶
solution, ∆𝑇 (°𝐶)
2. Heat change of water (J) -836 J
3. Heat change of salt (°𝐶) -78.5 J
4. Total enthalpy change (J) 1045 J
5. Δs (J/mol salt) 11346 J
6. Average Δs (J/mol salt)
Calculations:
A. Specific Heat of a Metal
Change in Temperature
39
38
37
TEMPERATURE (◦C)
36
35
34
33
32
31
30
0 50 100 150 200 250 300 350
TIME (S)
𝑀𝑎𝑠𝑠𝑤𝑎𝑡𝑒𝑟 = 𝑀𝑎𝑠𝑠𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑤𝑎𝑡𝑒𝑟 − 𝑀𝑎𝑠𝑠𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟
𝑀𝑎𝑠𝑠𝑤𝑎𝑡𝑒𝑟 = 49.65 𝑔 − 31.22 𝑔
𝑴𝒂𝒔𝒔𝒘𝒂𝒕𝒆𝒓 = 𝟏𝟖. 𝟒𝟑 𝒈
𝛥𝑇𝑤𝑎𝑡𝑒𝑟 = 𝑇𝐹𝑖𝑛𝑎𝑙 − 𝑇𝐼𝑛𝑖𝑡𝑖𝑎𝑙
𝛥𝑇𝑤𝑎𝑡𝑒𝑟 = 36°𝐶 − 32°𝐶
𝜟𝑻𝒘𝒂𝒕𝒆𝒓 = 𝟒°𝑪
𝑄𝑤𝑎𝑡𝑒𝑟 = 𝑀𝑎𝑠𝑠𝑤𝑎𝑡𝑒𝑟 × 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡𝑤𝑎𝑡𝑒𝑟 × 𝛥𝑇𝑤𝑎𝑡𝑒𝑟
4.18 𝐽
𝑄𝑤𝑎𝑡𝑒𝑟 = 18.43 𝑔 × × 4°𝐶
𝑔 ∙ °𝐶
𝑸𝒘𝒂𝒕𝒆𝒓 = 𝟑𝟎𝟖. 𝟏𝟓 𝑱
𝛥𝑇𝑚𝑒𝑡𝑎𝑙 = 𝑇𝐹𝑖𝑛𝑎𝑙 − 𝑇𝐼𝑛𝑖𝑡𝑖𝑎𝑙
𝛥𝑇𝑚𝑒𝑡𝑎𝑙 = 36°𝐶 − 98°𝐶
𝜟𝑻𝒎𝒆𝒕𝒂𝒍 = −𝟔𝟐°𝑪
𝑄𝑤𝑎𝑡𝑒𝑟 (𝐺𝑎𝑖𝑛𝑒𝑑) = −𝑄𝑚𝑒𝑡𝑎𝑙 (𝐿𝑜𝑠𝑡)
𝑄𝑚𝑒𝑡𝑎𝑙 = 𝑀𝑎𝑠𝑠𝑚𝑒𝑡𝑎𝑙 × 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡𝑚𝑒𝑡𝑎𝑙 × 𝛥𝑇𝑚𝑒𝑡𝑎𝑙
𝑄𝑚𝑒𝑡𝑎𝑙
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡𝑚𝑒𝑡𝑎𝑙 =
𝑀𝑎𝑠𝑠𝑚𝑒𝑡𝑎𝑙 × 𝛥𝑇𝑚𝑒𝑡𝑎𝑙
−308.15 𝐽
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡𝑚𝑒𝑡𝑎𝑙 =
10 𝑔 × −62°𝐶
−308.15 𝐽
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡𝑚𝑒𝑡𝑎𝑙 =
10 𝑔 × −62°𝐶
𝑱
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝑯𝒆𝒂𝒕𝒎𝒆𝒕𝒂𝒍 = 𝟎. 𝟒𝟗𝟕𝟎
𝒈 °𝑪
B. Enthalpy (Heat) of Neutralization for an Acid-Base Reaction
𝑯𝑪𝒍 + 𝑵𝒂𝑶𝑯
Change in Temperature
40
38
TEMPERATURE (◦C)
36
34
32
30
28
26
0 50 100 150 200 250 300 350
TIME (S)
𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑓 𝑎𝑐𝑖𝑑 + 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑓 𝑁𝑎𝑂𝐻
𝐴𝑣𝑒. 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝. 𝑜𝑓 𝐴𝑐𝑖𝑑 𝑎𝑛𝑑 𝑁𝑎𝑂𝐻 =
2
30°𝐶 + 22°𝐶
𝐴𝑣𝑒. 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝. 𝑜𝑓 𝐴𝑐𝑖𝑑 𝑎𝑛𝑑 𝑁𝑎𝑂𝐻 =
2
𝑨𝒗𝒆. 𝑰𝒏𝒊𝒕𝒊𝒂𝒍 𝑻𝒆𝒎𝒑. 𝒐𝒇 𝑨𝒄𝒊𝒅 𝒂𝒏𝒅 𝑵𝒂𝑶𝑯 = 𝟐𝟔°𝑪
𝛥𝑇 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑒𝑚𝑝. − 𝐴𝑣𝑒. 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝. 𝑜𝑓 𝐴𝑐𝑖𝑑 𝑎𝑛𝑑 𝑁𝑎𝑂𝐻
𝛥𝑇 = 38°𝐶 − 26°𝐶
𝜟𝑻 = 𝟏𝟐°𝑪
𝑉𝑜𝑙𝑢𝑚𝑒𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 𝑉𝑜𝑙𝑢𝑚𝑒𝑎𝑐𝑖𝑑 − 𝑉𝑜𝑙𝑢𝑚𝑒𝑁𝑎𝑂𝐻
𝑉𝑜𝑙𝑢𝑚𝑒𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 50 𝑚𝐿 − 50 𝑚𝐿
𝑽𝒐𝒍𝒖𝒎𝒆𝒎𝒊𝒙𝒕𝒖𝒓𝒆 = 𝟏𝟎𝟎 𝒎𝑳
𝑀𝑎𝑠𝑠𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 𝑉𝑜𝑙𝑢𝑚𝑒𝑚𝑖𝑥𝑡𝑢𝑟𝑒 × 𝐷𝑒𝑛𝑠𝑖𝑡𝑦𝑚𝑖𝑥𝑡𝑢𝑟𝑒
𝑔
𝑀𝑎𝑠𝑠𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 100 𝑚𝐿 × 1.0
𝑚𝐿
𝑴𝒂𝒔𝒔𝒎𝒊𝒙𝒕𝒖𝒓𝒆 = 𝟏𝟎𝟎 𝒈
𝑄 = 𝑀𝑎𝑠𝑠𝑚𝑖𝑥𝑡𝑢𝑟𝑒 × 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡𝑚𝑖𝑥𝑡𝑢𝑟𝑒 × 𝛥𝑇𝑚𝑖𝑥𝑡𝑢𝑟𝑒
4.18 𝐽
𝑄 = 100 𝑔 × × 12°𝐶
𝑔 ∙ °𝐶
𝑸 = 𝟓𝟎𝟏𝟔 𝑱
1𝐿 0.9832 𝑚𝑜𝑙
𝑚𝑜𝑙𝑂𝐻− = 50 𝑚𝐿 × ×
1000 𝑚𝐿 𝐿
𝒎𝒐𝒍𝑶𝑯− = 𝟎. 𝟎𝟒𝟗𝟏𝟔 𝒎𝒐𝒍
𝐻𝐶𝑙 + 𝑁𝑎𝑂𝐻 → 𝑁𝑎𝐶𝑙 + 𝐻2 𝑂
1 𝑚𝑜𝑙𝐻2 𝑂
𝑚𝑜𝑙𝐻2 𝑂 = 0.04916 𝑚𝑜𝑙 ×
1 𝑚𝑜𝑙𝑁𝑎𝑂𝐻
𝒎𝒐𝒍𝑯𝟐𝑶 = 𝟎. 𝟎𝟒𝟗𝟏𝟔 𝒎𝒐𝒍
𝑄
𝛥𝐻 = −
𝑚𝑜𝑙𝐻2 𝑂
1 𝑘𝐽
5016 𝐽 ×
1000 𝐽
𝛥𝐻 = −
0.04916 𝑚𝑜𝑙𝐻2 𝑂
𝒌𝑱
𝜟𝑯 = −𝟏𝟎𝟐. 𝟎𝟑𝟒𝟏
𝒎𝒐𝒍𝑯𝟐𝑶
𝑯𝑵𝑶𝟑 + 𝑵𝒂𝑶𝑯
Change in Temperature
33
32
TEMPERATURE (◦C)
31
30
29
28
27
26
0 50 100 150 200 250 300 350
TIME (S)
𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑓 𝑎𝑐𝑖𝑑 + 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑓 𝑁𝑎𝑂𝐻
𝐴𝑣𝑒. 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝. 𝑜𝑓 𝐴𝑐𝑖𝑑 𝑎𝑛𝑑 𝑁𝑎𝑂𝐻 =
2
32°𝐶 + 22°𝐶
𝐴𝑣𝑒. 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝. 𝑜𝑓 𝐴𝑐𝑖𝑑 𝑎𝑛𝑑 𝑁𝑎𝑂𝐻 =
2
𝑨𝒗𝒆. 𝑰𝒏𝒊𝒕𝒊𝒂𝒍 𝑻𝒆𝒎𝒑. 𝒐𝒇 𝑨𝒄𝒊𝒅 𝒂𝒏𝒅 𝑵𝒂𝑶𝑯 = 𝟐𝟕°𝑪
𝛥𝑇 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑒𝑚𝑝. − 𝐴𝑣𝑒. 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑇𝑒𝑚𝑝. 𝑜𝑓 𝐴𝑐𝑖𝑑 𝑎𝑛𝑑 𝑁𝑎𝑂𝐻
𝛥𝑇 = 32°𝐶 − 27°𝐶
𝜟𝑻 = 𝟓°𝑪
𝑉𝑜𝑙𝑢𝑚𝑒𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 𝑉𝑜𝑙𝑢𝑚𝑒𝑎𝑐𝑖𝑑 − 𝑉𝑜𝑙𝑢𝑚𝑒𝑁𝑎𝑂𝐻
𝑉𝑜𝑙𝑢𝑚𝑒𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 50 𝑚𝐿 − 50 𝑚𝐿
𝑽𝒐𝒍𝒖𝒎𝒆𝒎𝒊𝒙𝒕𝒖𝒓𝒆 = 𝟏𝟎𝟎 𝒎𝑳
𝑀𝑎𝑠𝑠𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 𝑉𝑜𝑙𝑢𝑚𝑒𝑚𝑖𝑥𝑡𝑢𝑟𝑒 × 𝐷𝑒𝑛𝑠𝑖𝑡𝑦𝑚𝑖𝑥𝑡𝑢𝑟𝑒
𝑔
𝑀𝑎𝑠𝑠𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = 100 𝑚𝐿 × 1.0
𝑚𝐿
𝑴𝒂𝒔𝒔𝒎𝒊𝒙𝒕𝒖𝒓𝒆 = 𝟏𝟎𝟎 𝒈
𝑄 = 𝑀𝑎𝑠𝑠𝑚𝑖𝑥𝑡𝑢𝑟𝑒 × 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐻𝑒𝑎𝑡𝑚𝑖𝑥𝑡𝑢𝑟𝑒 × 𝛥𝑇𝑚𝑖𝑥𝑡𝑢𝑟𝑒
4.18 𝐽
𝑄 = 100 𝑔 × × 5°𝐶
𝑔 ∙ °𝐶
𝑸 = 𝟐𝟎𝟗𝟎 𝑱
1𝐿 0.9832 𝑚𝑜𝑙
𝑚𝑜𝑙𝑂𝐻− = 50 𝑚𝐿 × ×
1000 𝑚𝐿 𝐿
𝒎𝒐𝒍𝑶𝑯− = 𝟎. 𝟎𝟒𝟗𝟏𝟔 𝒎𝒐𝒍
𝐻𝐶𝑙 + 𝑁𝑎𝑂𝐻 → 𝑁𝑎𝐶𝑙 + 𝐻2 𝑂
1 𝑚𝑜𝑙𝐻2 𝑂
𝑚𝑜𝑙𝐻2 𝑂 = 0.04916 𝑚𝑜𝑙 ×
1 𝑚𝑜𝑙𝑁𝑎𝑂𝐻
𝒎𝒐𝒍𝑯𝟐𝑶 = 𝟎. 𝟎𝟒𝟗𝟏𝟔 𝒎𝒐𝒍
𝑄
𝛥𝐻 = −
𝑚𝑜𝑙𝐻2 𝑂
1 𝑘𝐽
2090 𝐽 ×
1000 𝐽
𝛥𝐻 = −
0.04916 𝑚𝑜𝑙𝐻2 𝑂
𝒌𝑱
𝜟𝑯 = −𝟒𝟐. 𝟓𝟏𝟒𝟐
𝒎𝒐𝒍𝑯𝟐𝑶
C. Enthalpy (Heat) of Neutralization for the Dissolution of Salt
Change in Temperature
35
CHANGE IN TEMPREATURE (◦C)
30
25
20
15
10
0
0 50 100 150 200 250 300 350
TIME (S)
𝛥𝑇 = 𝑓𝑖𝑛𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 − 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
∆𝑇 = 20°𝐶 − 30°𝐶
∆𝑇 = −10°𝐶
𝑄 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = −𝑚𝑐∆𝑇
𝐽
𝑄 = (20𝑔) (4.18 ) (−10℃)
𝑔∙℃
𝑄 = −836 𝐽
𝑄 𝑜𝑓 𝑠𝑎𝑙𝑡 = 𝑚𝑐∆𝑇
𝐽
𝑄 = (5𝑔)(1.57 )(−10℃)
𝑔∙℃
𝑄 = −78.5 𝐽
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 = −(𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑎𝑙𝑡 + 𝑤𝑎𝑡𝑒𝑟)(𝑐 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟)(∆𝑇)
𝐽
∆𝐻 = −(25𝑔) (4.18 ) (−10℃)
𝑔∙℃
∆𝐻 = 1045 𝐽
−𝑄 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 + −𝑄𝑠𝑎𝑙𝑡
∆𝐻𝑠 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑎𝑙𝑡
−[(−836 𝐽) + (−78.5 𝐽)]
∆𝐻𝑠 =
0.0806 𝑚𝑜𝑙
𝐽
∆𝐻𝑠 = 11346
𝑚𝑜𝑙
