Question

At 1 atm, how much energy is required to heat 81.0 g of H2O(s) at –24.0 °C to H2O(g) at 141.0 °C?

=______ kJ

Answer 1

The total energy required is the sum of the energy to heat the -25.0 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 141 °C.

Step 1: Heat required to raise the temperature of ice from -25.0 °C to 141 °C Use the formula

q = mcΔT

where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature

q = (81 g)x(2.09 J/g·°C)[(0 °C - -24 °C)]
q = (81 g)x(2.09 J/g·°C)x(25 °C)
q = 4062.96 J

Heat required to raise the temperature of ice from -24 °C to 0 °C = 4232.5 J

Step 2: Heat required to convert 0 °C ice to 0 °C water

Use the formula

q = m·ΔHf

where
q = heat energy
m = mass
ΔHf = heat of fusion

q = (81 g)x(334 J/g)
q = 27054 J

Heat required to convert 0 °C ice to 0 °C water = 27054 J

Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water

q = mcΔT

q = (81 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]
q = (81 g)x(4.18 J/g·°C)x(100 °C)
q = 33858 J

Heat required to raise the temperature of 0 °C water to 100 °C water = 33858 J

Step 4: Heat required to convert 100 °C water to 100 °C steam

q = m·ΔHv

where
q = heat energy
m = mass
ΔHv = heat of vaporization

q = (81 g)x(2257 J/g)
q = 182817 J

Heat required to convert 100 °C water to 100 °C steam = 182817 J

Step 5: Heat required to convert 100 °C steam to 141 °C steam

q = mcΔT
q = (81 g)x(2.09 J/g·°C)[(141 °C - 100 °C)]
q = (81 g)x(2.09 J/g·°C)x(41 °C)
q =6940.89 J

Heat required to convert 100 °C steam to 150 °C steam = 6940.89 J

Step 6: Find total heat energy

HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4 + HeatStep 5

HeatTotal = 254732.85 J or 254.73 KJ