Heat Changes Involving Calories:
In a simple calorimeter all the heat released
from a chemical reaction goes into a measured quantity of water or
solution. It takes one calorie of heat to raise the temperature of 1 g of
water (or solution) one degree Celsius.
(1 g water = 1 cm3 H2O
= 1 mL H2O).
Example:
Calculate the heat released per gram of a
substance if 10 g of it changed the temperature of 150 mL of water from 20.0oC
to 25.0oC.
Solution:
150 g H2O x 5.00 oC (temperature
change) x 1 cal / 1 g x 1oC =
750 calories released
750 cal = 75 cal
10 gram 1 gram
Specific Heat Values - Joules
| Specific Heat |
J / g oC |
| Water (liquid) |
4.18 |
| Water (gas) |
1.87 |
| Water (solid) |
2.077 |
| Ethanol, C2H5OH(l) |
2.438 |
| Methane, CH4(g) |
2.200 |
| Isooctane, C8H18(l) |
2.093 |
| Aluminum, Al(s) |
0.897 |
| Table salt, NaCl(s) |
0.865 |
| Graphite, C(s) |
0.714 |
| Iron, Fe(s) |
0.449 |
| Silver, Ag(s) |
0.235 |
| Mercury, Hg(l) |
0.139 |
| Tungsten, W |
0.132 |
| Lead, Pb(s) |
0.129 |
| Gold, Au(s) |
0.129 |
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Heating
Curve of Water
Calorimetry
1 WS Calorimetry
2 WS
Causes
of Change Outline
Example 1: In going from ice at -34oC
to steam at 138oC, a sample of water absorbs 1.41 x 105
J. Find the mass of the sample.
This problem will involve 5 steps:
Use X for the mass of the sample in
all steps of the problem.
Recall: DT
= Tf - Ti
Step
1: Warm the ice from -34 oC to 0oC.
equation:
DH1
= (mass)(Cp)( DT)
Cp ice (s) = 2.077 J/g oC
substitute: DH1
= (X g)(2.077 J/g oC)(0
oC
- (-34 oC))
Step 2: Melt the
ice at 0oC to water at 0oC
equation:
DH2
= (mass)(Cf)
Cf water = 333
J/g
substitute: DH2
= (X g)(333 J/g)
Step 3: Warm the
water from 0oC to 100 oC
equation: DH3
= (mass)(Cp)( DT)
Cp water (l) = 4.184 J/g oC
substitute: DH3
= (X g)(4.184 J/g oC)(100
oC
- 0 oC)
Step 4: Vaporize the
water (boil) it from a liquid at 100oC to steam at 100oC
equation: DH4
= (mass)(Cv)
Cv water = 2256
J/g
substitute: DH4
= (X g)(2256 J/g)
Step 5: Superheat the
steam from 100oC to 138oC.
equation: DH5
= (mass)(Cp)( DT)
Cp water (g) = 1.87 J/g oC
substitute: DH5
= (X g)(1.87 J/g oC)(138
oC
- 100 oC)
Finally,
the
total heat absorbed by the sample will be equal to the sum of the heat absorbed
by each step in the process:
DH1
+
DH2
+
DH3
+
DH4
+
DH5
= 1.41 x 105
J
Substitute:
(X g)(2.077 J/g oC)(0
oC
- (-34 oC))
+ (X g)(333 J/g) + (X g)(4.184 J/g oC)(100
oC
- 0 oC)
+ (X g)(2256 J/g) + (X g)(1.87 J/g oC)(138
oC
- 100 oC)
Solve for X: X
= 44.7 g
Example 2: If 20 g
of silver at 350oC are mixed with 200 g of water at 30oC,
find the final temperature of the system.
This problem will involve several
steps.
Use X for the final temperature of the system. Recall:
DT =
Tf - Ti
The silver loses heat (energy = (-)) while
the water gains heat (energy = (+).
Step 1: -
DHsilver
= DHwater
(Energy 'lost' by silver is equal to heat 'gained'
by water)
equation:
- (mass)(Cp)(
DT)
= (mass)(Cp)(
DT)
Cp silver (s) = 0.235 J/g oC
Cp water (s) = 4.184 J/g oC
substitute: - (20
g)(0.235 J/g oC)(X
oC
- 350 oC)
= (200 g)(4.184 J/g oC)(X
oC
- 30 oC)
solve: - 4.7X + 1645 = 836X - 25,080
combine like terms: 26.725
= 840.7X
X = 31.8oC
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