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Water of mass 3kg initially at 20℃ is heated in an electric kettle rated 3.0kW. Thewater is heated until it boils at 100℃. Taking specific heat capacity of water=4200Jkg/K, heat capacity of kettle=450J/kg, specific latent heat of vapourisation of water =2.3MJ/kg, calculate;

  1. The heat absorbed by the water.
  2. Heat absorbed by the electric kettle.
  3. The time taken for the water to boil .
  4. How much longer it will take to boil away all the water? 

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  1. The heat absorbed by the water. 
    • Q = MCΔθ
      = 3kg x 4200JKg-1k-1 x 80 
      = 240 x 4200 = 1008000J / 1.008kJ 
  2. Heat absorbed by the electric kettle. 
    • Q = MkCkΔθ
      = 450 x 80  = 36000J
  3. The time taken for the water to boil . 
    • pt = total heat absorbed
      pt = 36000 + 10080001
      3000t = 1044000
      t = 1044000/3000
      t = 348 sec
  4. How much longer it will take to boil away all the water? (3 marks)
    • pt = mLv
      3000t = 3 x 2.3 x 106
      t = (3 ×2.3 × 106)/3000
      t = 2300 seconds
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