An electric heater used to boil small amounts of water consists of a 25-Ω coil that is immersed directly in the water. It operates from a 60-V socket. How much time is required for this heater to raise the temperature of 0.50 kg of water from 29° C to the normal boiling point?

Answer 1

i = V/R = 60/25 = 2.4 amps
P = i v = 2.4*60 = 144 watts or Joules/second

Joules into water = 144 t
where t is in seconds

144 t = C (.5)(100-29)
I do not recall C, specific heat of water
in Joules/ kilogram deg centigrade

Answer 2

t(seconds) = ((kg water)(temp change)(4186))/Watts

Watts = V^2/ohms