How much heat is required to convert 84.9 g of ice at -17.0 C to steam at 114 C?
Q = heat change for conversion of ice at -17 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 114 oC
Amount of heat absorbed , Q = mcdt + mL + mc'dt + mL' + mc"dt"
= m(cdt + L + c'dt' + L' + c"dt" )
m = mass of ice = 84.9 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’’ = 114-100 = 14oC
dt' = 100 -0 =100 oC
dt = 0-(-17)=17 oC
Plug the values we get Q = m(cdt + L + c'dt '+ L' + c"dt" )
= 261.3x103 J
= 261.3 kJ