Question

At 1 atm, how much energy is required to heat 93.0 g of H2O(s) at –22.0 °C to H2O(g) at 171.0 °C?

Answer 1

specific heat capacity of water, C1 = 4.186 J/goC
specific heat capacity of ice, C2 = 2.09 J/goC
specific heat capacity of steam, C3 = 2.01 J/goC
Latent heat of fusion of ice, Lf = 334 J/g
latent heat of vaporization of water LV = 2264.76 J/g
m= 93 g

heat required to take ice from -22 oC to 0 oC,
Q1 = m*C2*delta T
      =93*2.09*(22)
      = 4276.14 J

heat required to convert ice to water,
Q2 = m* Lf
      = 93*334
      = 31062 J

heat required to take water from 0 oC to 100 oC,
Q3 = m*C1*delta T
      =93*4.186*(100-0)
      = 38929.8 J

heat required to convert water to steam,
Q4 = m* LV
      = 93*2264.76
      = 210622.68 J

heat required to take steam from 100 oC to 171 oC,
Q5 = m*C3*delta T
      =93*2.01*(171-100)
      = 13272.03 J

Total heat required= Q1 + Q2 + Q3 + Q4 + Q5
      = 4276.14 + 31062 + 38929.8 + 210622.68+ 13272.03
     = 298163 J
     = 298 KJ
Answer: 298 KJ