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Measuring Quantity of Heat
Module: Water
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Heat Absorbed By The Environment
1
Calculate the amount of heat required to heat 155g from 15.0o
C to33.0oC.
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Qwater =mwaterCT
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Qwater =mwaterCT
The sign of the quantity of heat is positive since the sign is dependant onT
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Qwater =mwaterCT
The sign of the quantity of heat is positive since the sign is dependant onT
Tf>Ti
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Qwater =mwaterT (1)
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Qwater =mwaterCT
= (155g)(4.18J/g/oC)(33.0oC 15.0oC)
= 1.17 104J
(2)
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Molar Heat of Solution
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Molar Heat of Solution
The Molar Heat of Solution (Hsoln) of a substance is the amount of
heat energy absorbed when one mole of a substance dissolves inexcess water.
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Molar Heat of Solution
The Molar Heat of Solution (Hsoln) of a substance is the amount ofheat energy absorbed when one mole of a substance dissolves in
excess water.Remember:
Hsoln =HProducts HReactants
Where by convention HProducts
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Molar Heat of Solution
The Molar Heat of Solution (Hsoln) of a substance is the amount ofheat energy absorbed when one mole of a substance dissolves inexcess water.
Remember:Hsoln =HProducts HReactants
Where by convention HProducts
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Molar Heat of Solution
The Molar Heat of Solution (Hsoln) of a substance is the amount ofheat energy absorbed when one mole of a substance dissolves inexcess water.
Remember:Hsoln =HProducts HReactants
Where by convention HProducts
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Process Of Calculating The Molar Heat of Solution
We assume the dissolution reaction to be exothermic. As an equation:
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Process Of Calculating The Molar Heat of Solution
We assume the dissolution reaction to be exothermic. As an equation:
Qsolution = mCT
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Process Of Calculating The Molar Heat of Solution
We assume the dissolution reaction to be exothermic. As an equation:
Qsoln =
msolnCT
The molarheat of solution is then calculated as the energy released permoleof the substance
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Process Of Calculating The Molar Heat of Solution
We assume the dissolution reaction to be exothermic. As an equation:
Qsoln = msolnCT
The molarheat of solution is then calculated as the energy released permoleof the substance:
Hsoln = Qsoln
nsubstance
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E l Q i
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Example Question
A calorimeter contained 50.0g of water. When 1.0g of dry ammonium
chloride powder was dissolved in water, the mixture dropped intemperature from 21.3oC to 21.3oC.
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E l Q i
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Example Question
A calorimeter contained 50.0g of water. When 1.0g of dry ammoniumchloride powder was dissolved in water, the mixture dropped intemperature from 21.3oC to 20.1oC.
Assume that no energy was lost to the environment, calculate the molarenthalpy of solution for this dissolution reaction (Assume C = 4.2J/g/oC)
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E l Q i
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Example Question
Data:
mwater = 50.0g
mNH4Cl= 1.0g
T = 20.1oC 21.3oC = 1.2oC4.2J/g/oC
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E l Q ti
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Example Question
Qsoln = msolnCT
=
(mwater+ mNH4Cl)CT= (51.0g)(4.2J/g/oC)(1.2oC)
= 257.04J
(3)
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E l Q ti
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Example Question
Hsoln = Qsoln
n(NH4Cl)
= 257.04J1.0g
53.492mol
= +13753J/mol
(4)
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Proper Understanding Of The Formula
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Proper Understanding Of The Formula
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Proper Understanding Of The Formula
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Proper Understanding Of The Formula
Consider the fact that no heat is lost during the dissolution process. i.e.
Qlost+ Qgained= 0
Qreaction+ Qcalorimeter = 0
We note that reaction refers to the dissolution of the salt (which is still areaction)
Qreaction= Qcalorimeter
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Proper Understanding Of The Formula
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Proper Understanding Of The Formula
Qreaction= Qcalorimeter
Now, Qcalorimeterrefers to the heat absorbed by the calorimeter.
Qreaction = mcalCT
But, mcal refers to the mass of the solution.
Qreaction= msolutionCT
And Qsolution =Qreaction
Qsolution = msolutionCT
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