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30 April, 23:56

10. What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its

temperature changes from 25°C to 20°C?

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Answers (1)
  1. 1 May, 01:30
    0
    c = 4016.64 j/g.°C

    Explanation:

    Given dа ta:

    Mass of substance = 2.50 g

    Calories release = 12 cal (12 * 4184 = 50208 j)

    Initial temperature = 25°C

    Final temperature = 20°C

    Specific heat of substance = ?

    Formula:

    Q = m. c. ΔT

    Q = amount of heat absorbed or released

    m = mass of given substance

    c = specific heat capacity of substance

    ΔT = change in temperature

    Solution:

    Q = m. c. ΔT

    ΔT = T2 - T1

    ΔT = 20°C - 25°C

    ΔT = - 5°C

    50208 j = 2.50 g. c. - 5°C

    50208 j = - 12.5 g.°C. c

    50208 j / -12.5 g.°C = c

    c = 4016.64 j/g.°C
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