Question

Please solve question number 10 with detailed explanation.
Answer is also given below the question in the above given picture.

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