Asked by Dee
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?
Answers
Answered by
Damon
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
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
Answered by
Anonymous
t(seconds) = ((kg water)(temp change)(4186))/Watts
Watts = V^2/ohms
Watts = V^2/ohms
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