A Shiny electric kettle with a 2.0kw heating element has a thermal capacity of 400j/k. 1kg of water at 20degrees celsius is placed in the kettle. The kettle is then switched on and 0.5kg of water remain after 13mins. Ignoring heat loss, calculate the value of specific heat of evaporation

Answer 1

2KW is 7200 kJ in one hour.
So how much does it take to raise temperature of the kettle to 100 C.
400 x (Tfinal-Tinitial) = ?
How much to heat 1000 g H2O from 20 to 100.
q in kJ = 1 kg x 4.18 x (Tfinal-Tinitial) = ?
Add heat capacity q to q to raise 1000 g H2O to find total heat required to raise kettle and water to boiling.
How much heat do you have in that 13 min.
7200 x 13/60 = ? kJ
How much heat is left after getting H2O to the boiling point. That's
total J in 13 min - heat needed to raise kettle and water to 100 C = ?
Then heat vaporization x 500 g = heat left to boil water. The value in text is 40.65/mol. This is close to that but about 8% high.