Physics Question (An electric heater is used to boil...)?

An electric heater is used to boil small amounts of water and consists of a 15 ohm coil that is immersed directly in water. It operates from a 12.0V socket. How much time is required for this heater to raise the temperature of 0.50KG water from 13deg C to the normal boiling point?











I didn't understand this problem at all.. the book states that it is a harder question, and I do not know how to get the answer.





V= MC(DeltaT)/Q





Is that the equation that I use?








can anyone help me?





thanks


-aly

Physics Question (An electric heater is used to boil...)?
energy required to heat water from 13 to 100 deg C





Q = m c delta T = 0.5 (kg) 4181(I/kg-oC) (100-13)


Q = 181873.5 Joule (c sp heat of h2o)





This Q will have ro be supplied by electric heater





Power supplied by heater = V^2 /R = Q / time





Q = V^2 *t / R


t = R Q /V^2 = 15*181873.5 /12*12 = 18945.16 secs


t = 5.262 hours





time is high because voltage is very low
Reply:Volts (V) = Current (A) x Resistance (R)


A = V/R = 12/15 = 0.8Amps.


Watts = V x A = 12 x 0.8 = 9.6 Watts


1Joule/second = 1 Watt. = 9.6J/s





To heat the water, requires 500g x 87°C(ΔT) x 4.184J/g/°C


= 182,000J


182,000 ÷ 9.6J/s = 18,959seconds


= 316minutes. = 5.3 hours.
Reply:the power of the heater is v^2/R


= 12^2/15 = 9.6 watts.


This means 9.6 jouls of heat per second.


The heat required by water


= mass of water * specific heat * temp rise


=0.5*4185*(100-13)


note- specific heat of water = 4185 j/kg/deg k and boiling point of water =100)


so heat required = 182047 J


time required = 182047/9.6 = 18963 seconds.