Using the following data for water, determine the energy required to heat 27.0 g
ID: 1027844 • Letter: U
Question
Using the following data for water, determine the energy required to heat 27.0 g of ice (solid water) at – 12.0oC through water and vapor at 120 oC..
Boiling point
373 K
Melting point
273 K
Enthalpy of vaporization
2,260 J/g
Enthalpy of fusion
334 J/g
Specific heat capacity (solid)
2.11 J/(g K)
Specific heat capacity (liquid)
4.18 J/(g K)
Specific heat capacity (gas)
2.08 J/(g K)
Boiling point
373 K
Melting point
273 K
Enthalpy of vaporization
2,260 J/g
Enthalpy of fusion
334 J/g
Specific heat capacity (solid)
2.11 J/(g K)
Specific heat capacity (liquid)
4.18 J/(g K)
Specific heat capacity (gas)
2.08 J/(g K)
Explanation / Answer
Q = heat change for conversion of ice at -12 oC to ice at 0 oC + heat change for
conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 100 oC +heat change for conversion of water at 100 oC to vapour at 100 oC+ heat change for conversion of vapour at 100 oC to vapour at 120 oC
Amount of heat absorbed , Q = mcdt + mL + mc'dt + mL' + mc"dt"
= m(cdt + L + c'dt' + L' + c"dt" )
Where
m = mass of water = 27.0 g
c” = Specific heat of steam = 2.11 J/g degree C
c' = Specific heat of water = 4.18 J/g degree C
c = Specific heat of ice= 2.08 J/g degree C
L’ = Heat of Vaporization of water = 2260 J/g
L= Heat of fusion of ice = 334 J/g
dt’’ = 120-100 = 20oC
dt' = 100 -0 =100 oC
dt = 0-(-12)=12 oC
Plug the values we get Q = m(cdt + L + c'dt '+ L' + c"dt" )
Q= 83.14*10^3 J
Q= 83.14 kJ
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