Question 377195
How much energy is required to heat 1.5 L of water at a temperature of 20 °C to the boiling point 100 ºC?
 Sol'n:
  Q = mCp(T2-T1)
  m = mass density of water = 1 kg/L for Water, thus m = 1.5-kg
  Cp = 4.186 kJ/kg-ºC, T2 = 100ºC & T1 = 20ºC
  Then, 
    Q = 1.5(4.186)(100-20)
    Q = 502.32 kJ ----answer

There is 1.5 L of water in a kettle on a cooking plate of 2.0 KW. It takes 7 minutes to heat the water from 20 C to the poiling point (100 C). How much electrical energy is required ? and what is the efficiency of the heating process?
 Sol'n:
 From above: Q = 502.32-kJ, energy needed to raise the temp. of water.
 From: Q = (eff.)E
 where: E - the energy needed by the kettle
       eff.= efficiency of the kettle
     E = Pt
       = 2,0000 J/s x 7(60 s.)     
     E = 840-kJ ----answer
thus,
  eff. = Q/E (100)
       = 502.32/840 (100)
  eff. = 59.8% ----answer

just send your question to my email:
rfadrogane@rocketmail.com