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16 February, 23:17

When 28 grams of a metal are heated from 14.4°C to 150°C it takes 23,200). What is the specific heat of the metal?

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Answers (2)
  1. 16 February, 23:45
    0
    6.11 J/g°C

    Explanation:

    q = mc (delta theta)

    23200 = 28 * c * (150 - 14.4)

    23200 = 28 * c * 135.6

    c = 23200 / (28 * 135.6)

    c = 6.110408765
  2. 16 February, 23:46
    0
    6.11 J/g°C

    Explanation:

    We need to use the heat equation: q = mCΔT, where q is the amount of heat required in Joules, m is the mass in grams, C is the heat capacity, and ΔT is the change in temperature.

    Here, we know q = 23,200 Joules, m = 28 grams, and the ΔT is just the difference of the temperatures: 150 - 14.4 = 135.6°C. Substitute these values to find C:

    q = mCΔT

    23,200 = 28 * C * 135.6

    C = 6.11

    Thus, the specific heat is 6.11 J/g°C.
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