#8 please and show work. thank you. Changes of Phase Problems: For each of the l
ID: 496702 • Letter: #
Question
#8 please and show work. thank you.
Explanation / Answer
The amount of heat lost ,
Q = heat change for conversion of vapour at 150 oC to vapour at 100 oC + heat change for conversion of steam at 100 oC to water at 100 oC + heat change for conversion of water at 100oC to water at 0 oC + heat change for conversion of water at 0oC to ice at 0oC + heat change for conversion of ice at 0 oC to ice at -15 oC
Amount of heat released , Q = mcdt + mL + mc'dt + mL' + mc"dt"
= m(cdt + L + c'dt' + L' + c"dt" )
Where
m = mass of steam = 120.0 g
c = Specific heat of steam = 2.1 J/g degree C
c' = Specific heat of water = 4.186 J/g degree C
c" = Specific heat of ice= 2.09 J/g degree C
L = Heat of Vaporization of water = 2260 J/g
L'= Heat of fusion of ice = 334.9 J/g
dt = 150-100 = 50oC
dt' = 100 -0 =100 oC
dt'' = 0-(-15) = 15 oC
Plug the values we get Q = m(cdt + L + c'dt '+ L' + c"dt" )
= 378.0x103 J
= 378.0 kJ
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