How much energy is required to convert 100.0 g of water at 20.0 °C completely to steam at 100.0 °C?

1 Answer
Jul 25, 2014

Answer:

To convert 100.0 g of water at 20.0 °C to steam at 100.0 °C requires 259.5 kJ of energy.

Explanation:

This is like the Socratic problem here.

For this problem, there are only two heats to consider:

#q_1# = heat required to warm the water from 20.0 °C to 100.0 °C.
#q_2# = heat required to vapourize the water to steam at 100 °C.

#q_1 = mcΔT = "100.0 g × 4.184 J"^°"C"^(-1)"g"^(-1) × 80.0^°"C" = "33 472 J" #

#q_2 = mΔH_"vap" = "100.0 g × 2260 J·g"^(-1) = "226 000 J"#

#q_1 + q_2 = "( 33 472 + 226 000) J = 259 472 J = 259.5 kJ"# (4 significant figures)