Question

Please solve Q. no. 10 and 11 with detailed explanations.
Answers are also given with the questions.

Answer 1

10. a.  Specific heat of water = s = 4200 J/kg /°C

   It means that 1 kg of water requires /releases 4200 J of heat , when heated or cooled by 1°C.
  So answer is 4, 200 Joules

  b. Amount of heat released by 0.02 kg during the cooling from 70°C to 45°C
         = m s ΔT   , m = 0.02 kg,   s = 4200 J/kg/°C ,  ΔT = (70-45) = 25°C
         = 0.02 * 4200 * 25 = 2,100 J

   c) Temperature of calorimeter increased by ΔT = 45 - 15 = 30°C
        mass m = 0.16 kg
        Heat absorbed = H = 2, 100 J
        H = m s ΔT
        2100 = 0.16 * s * 30
        s = 7000/16 = 437.5 J /kg/°C
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Q11. 

    Heat lost by the metal piece = m s ΔT 
       = 20 g * 0.3 J/g/°C * (100° - 22°)C = 20 * 0.3 * 78 J
       = 468 J

    Heat gained by the calorimeter and water = 468 J
    468J = Heat gained by calorimeter + heat gained by water
            = m1 s1 ΔT1 + m2 s2 ΔT2
           = 50 g * 0.42 J/g/°C * (22-20)°C +  m2 * 4.2 J/g/°C * (22 - 20)°C
           = 42 J + 8.4 m2   Joules
    m2 = 426/8.4 = 50.71 gms