Heat of Vaporization and Heat of Fusion Learning Goal: To calculate the total en
ID: 1063526 • Letter: H
Question
Heat of Vaporization and Heat of Fusion
Learning Goal:
To calculate the total energy required to convert a solid to a gas using heat capacity, heat of fusion, and heat of vaporization.
For example, the heat of fusion for H2O is 6.02 kJ/mol, which means that 6.02 kJ of heat is needed to melt a mole of ice. Twice as much heat is needed to melt twice as much ice, as shown here:
2.00 mol×6.02 kJ/mol=12.0 kJ
Constants
Heat of fusion (Hfus) is used for calculations involving a phase change between solid and liquid, with no temperature change. For H2O, Hfus=6.02 kJ/mol.
Specific heat capacity (C) is used for calculations that involve a temperature change, but no phase change. For liquid water, C=4.184 J/(gC).
Heat of vaporization (Hvap) is used for calculations involving a phase change between liquid and gas, with no temperature change. For H2O, Hvap=40.7 kJ/mol.
Part A
How much heat is required to melt 52.5 g of ice at its melting point?
Express your answer numerically in kilojoules.
17.5
SubmitHintsMy AnswersGive UpReview Part
Correct
Part B
How much heat is required to raise the temperature of 52.5 g of water from its melting point to its boiling point?
Express your answer numerically in kilojoules.
158.0775
SubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 4 attempts remaining; no points deducted
****NEED HELP PLEASE PART B ANSWER IS WRONG****
Heat of Vaporization and Heat of Fusion
Learning Goal:
To calculate the total energy required to convert a solid to a gas using heat capacity, heat of fusion, and heat of vaporization.
The amount of heat required to melt one mole of a solid is called the heat of fusion (Hfus). The amount of heat required to vaporize (boil) one mole of a liquid is called the heat of vaporization (Hvap). The heat of fusion and the heat of vaporization can be used to calculate the amount of heat needed to melt or vaporize, respectively, a given amount of substance.For example, the heat of fusion for H2O is 6.02 kJ/mol, which means that 6.02 kJ of heat is needed to melt a mole of ice. Twice as much heat is needed to melt twice as much ice, as shown here:
2.00 mol×6.02 kJ/mol=12.0 kJ
Constants
Heat of fusion (Hfus) is used for calculations involving a phase change between solid and liquid, with no temperature change. For H2O, Hfus=6.02 kJ/mol.
Specific heat capacity (C) is used for calculations that involve a temperature change, but no phase change. For liquid water, C=4.184 J/(gC).
Heat of vaporization (Hvap) is used for calculations involving a phase change between liquid and gas, with no temperature change. For H2O, Hvap=40.7 kJ/mol.
Part A
How much heat is required to melt 52.5 g of ice at its melting point?
Express your answer numerically in kilojoules.
17.5
kJSubmitHintsMy AnswersGive UpReview Part
Correct
Part B
How much heat is required to raise the temperature of 52.5 g of water from its melting point to its boiling point?
Express your answer numerically in kilojoules.
158.0775
kJSubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 4 attempts remaining; no points deducted
****NEED HELP PLEASE PART B ANSWER IS WRONG****
Explanation / Answer
q = m c dT -- equation 1
Where, q = heat change , m= mass in gram,
c = specific heat (in terms of J/gK)
dT = final temperature – initial temperature.
Part B. Given,
Mass of water = 52.5 g
Initial temperature of water = 00C [Melting point]
Final temperature of water = 1000C [Boiling point]
dT = 1000C – 00C = 1000C
Specific heat = 4.184 J/g0C
Putting the values in equation 1
q = 52.5 g x (4.184 J/g0C) x 1000C
or, q = 21966 J = 21.966 kJ ; [1 kJ = 1000 J]
Therefore, the amount of heat required to raise the temperature = 21.966 kJ
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