11.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    11.14 Mass of the metal, m = 0.20 kg = 200 g

    Initial temperature of the metal,  T1 = 150 °C, Final temperature of the metal,  T2 = 40 °C

    The water equivalent mass of the calorimeter, m’ = 0.025 kg = 25 g

    Volume of water, V = 150 cm3

    Mass of water, M at T = 27 °C, = 150 ×1=150g

    Fall in metal temperature,  ?  T = T1-T2 = 150 – 40 = 110 °C

    Specific heat of water,  Cw = 4.186 J/g/ ° K

    Let the specific heat of metal = C

    Then, heat loss by the metal,  θ = mC ?  T ……. (i)

    Rise in the water of the calorimeter system ?  T’ = 40 – 27

    ...more

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