Answer 1:
This is a breadbox physics/engineering
question.
The process is a fairly simple one:
A
container
with some amount of water (lets say 1 kg for
simplicity) is heated using some source (let's
say an electric heating element that outputs
Q (energy)= 2,500 Watts or Joules/second)(Joules
is an energy unit).
All we need now is
the starting temperature of the water (we'll say
25 Centigrade) and the heat capacity (4186
Joules/kg*Kelvin)(Kelvin is a unit of
temperature degrees).
So, how much heat do we need to add to 1kg of
water in order to heat it from T_{0}=25
Centigrades to T_{1}=100 Centigrade?
Well,
this is precisely what heat capacity means, it's
been measured, and it is ~4186 Joules/kg
Kelvin.
Q_Required = (100  25) Kelvin * 4186 Joules/
(kg Kelvin) * 1kg
= 314,000 Joules
We said earlier that our stove supplies ~
2,500
Joules/second, so 314,000 Joules / (2,500
Joules/second) = 126 seconds.
Time it! Remember 1 kg is about 1
Liter of
water. Our estimate is likely low (based on
personal experience) and is likely due to an
over estimate of the heat output of the stove
(not 100% of the stove's heat goes directly to
the water; much of it is wasted in hot air
escaping around the sides of the kettle).
Click Here to return to the search form.
