Answer
45.36372 kilojoules
Step-by-step explanation
Given:
Mass of H2O = 20.0 g
Initial temperature, T₁ = 20 °C
Final temperature, T₂ = 100 °C
Change in temperature, ΔT = T₂ - T₁ = 100 °C - 20 °C = 80 °C
Specific heat of water, c = 4.186 J/g°C
What to find:
The kilojoules required to change the temperature of 20.0 g of H2O from 20 Celsius to steam at 100 Celsius.
Step-by-step solution:
For this problem, there are only two heats to consider:
q₁ = heat required to warm the water from 20.0 °C to 100.0 °C.
q₂ = heat required to vapourize the water to steam at 100 °C.
q₁ = mcΔT = 20.0 g x 4.186 J/g°C x 80 °C = 163.72 joules
q₂ = mΔHvap = 20.0 g x 2260J/g = 45200 joules
Therefore, q₁ + q₂ = (163.72 + 45200)J = 45363.72 J = 45.36372 kJ.
Thus, 45.36372 kilojoules are required to change the temperature of 20.0 g of H2O from 20 Celsius to steam at 100 Celsius.