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physical behavior of matter.....aim 11 |
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how is the heat absorbed or released calculated? |
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aim 12..... |
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The heat of fusion is the heat needed to melt 1 gram of a solid at constant T. The quantity of heat that is absorbed during fusion or melting is released during solidification or freezing.
q = mHf
m = mass
Hf = heat of fusion
Example
The heat of fusion of water is 334 J/g See table B. Therefore, to melt 1 g of ice at 0 oC 334 J are required .
Problems
1. What is the heat needed to melt 100.0 grams of ice?
2. How many grams of ice can be melted with 668 J?
3. What is the of fusion of substance X if 200.0 J are required to melt 10.0 g of X?
4. What is the heat released when 2.0 kg of water freezes?
The heat of vaporization is the heat needed to vaporize 1 gram of a liquid at constant T. The quantity of heat that is absorbed during vaporization is released during condensation.
Heat: q = mHv
m = mass
Hv = heat of vaporization
Example
The heat of vaporization of water is 2260 J/g See table B. Therefore, to vaporize 1 g of water at 100 oC 2260 J are required .
Problems
1. What is the heat needed to vaporize 1000.0 grams of water?
2. How many grams of water can be vaporized with 6780 J?
3. What is the of vaporization of substance Y if 500.0 J are required to vaporize 100.0 g of X?
4. What is the heat released when 2000.0 mg of water condenses?
The heat involved in temperature change for a given sample of matter can be calculated using a formula.
q = mCΔT
q = heat
m = mass of sample
C = specific heat of substance. See table B for specific heat of water (4.18 J/g.oC)
The specific heat of a substance is the amount of energy in Joules (J) required to raise 1 gram (g) of the substance by 1 degree Celcius (oC)
Note: The difference between 1oC and 1K is the same.
ΔT = change of temperature (Tfinal - Tinitial)
A calorimeter is an instrument for measuring heat absorbed or released during reactions.

Example
1. A sample of compound X was burned in a bomb calorimeter containing 100.0 g of water. The initial temperature of the water was 25oC and it increased to 80oC. How many joules were released?
q =?
m = 100. g
C = 4.18 J/g.oC from table B
Tinitial = 25oC
Tfinal = 80oC
ΔT = Tfinal - Tinitial = 80oC - 25oC = 55oC
q = (100.0 g) (4.18 J/g.oC) (55 oC) = 23,100J
2. Using the diagram below representing a crude calorimeter, calculate the amount of heat absorbed by the water.
q = ?
m = 30 g
C = 4.2 J/g.K
ΔT = 37.5oC - 25oC = 12.5oC or 12.5 K
q = (30 g) (4.2 J/g.K) (12.5 K) = 1575 J

Problems
1.How many joules are absorbed when the temperature of 100.0 grams of water changes from 300 K to 350 K?
2. How many grams of H2O(l) absorb 400.0 J when temperature increases by 10 K?
3. What is the specific heat of substance X if 300.0 J are absorbed by 200.0 g of X as temperature change from 290 K to 300 K?
4. What is the change in temperature of 1.0 kg of H2O(l) when 400.0 J are released?
5. What is the initial temperature of 30.0 g of H2O(l) that released 2.0 kJ as the temperature dropped to 10oC?
TEST YOUR UNDERSTANDING
6/03
Calculate the heat released when 25.0 grams of water freezes at 0oC. Show all work. Record your answer with an appropriate unit.
8/04
42 How much heat energy must be absorbed to completely melt 35.0 grams of H2O(s) at 0°C?
(1) 9.54 J (2) 146 J (3) 11 700 J (4) 79 100 J
1/05
Base your answers to questions 70 through 71 on the information below.
In a laboratory experiment, 10.00 grams of an unknown solid is added to 100.0 milliliters of water and the temperature of the resulting solution is measured over several minutes, as recorded in the table below.

70 On a graph, plot the data from the data table. Circle and connect the points.
71 Given the statement: The unknown solid is either sodium hydroxide or lithium bromide, and both of these compounds dissolve in water exothermically.
a Explain how the experimental data support the statement.
b State specific information from Reference Table I to support the statement.
Base your answers to questions 73 through 76 on the information below.




Figure 1: In a laboratory investigation, a 50.0-gram sample of copper is at 100.0°C in a boiling water bath.
Figure 2: A Styrofoam cup with a lid is used as a calorimeter. The cup contains 100.0 grams of distilled water at 23.2°C.
Figure 3: The hot copper is poured into the cup of water, and the cup is quickly covered with the lid.
Figure 3: A thermometer is inserted through the lid. The copper and water are gently stirred in the cup. The temperature is checked periodically. The highest temperature noted is 26.3°C.
73 In terms of energy flow, explain why the temperature of the water in the calorimeter increases.
74 Using
the information given, complete the data table below.
75 Show a correct numerical setup for calculating the number of joules of heat gained by the water.
76 In this investigation, the change in heat of the copper is greater than the change in heat of the water. What error could account for this apparent violation of the Law of Conservation of Energy? Do not use human error as part of the answer.
6/05
Base your answers to questions 67 through 70 on the information below. A substance is a solid at 15°C.
A student heated a sample of the solid substance and recorded the temperature at one-minute intervals in the data table below.

67 On a graph, plot the data from the data table. Circle and connect the points.
68 Based on the data table, what is the melting point of this substance?
69 What is the evidence that the average kinetic energy of the particles of this substance is increasing during the first three minutes?
70 The heat of fusion for this substance is 122 joules per gram. How many joules of heat are needed to melt 7.50 grams of this substance at its
melting point?