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20 June, 08:01

Addition of 50. J to a 10.0-g sample of a metal will cause the temperature of a metal to rise from 25ºC to 35ºC. The specific heat of the metal is closest to0.0005 J / (g·ºC) 0.50 J / (g·ºC) 2.5 J / (g·ºC) 4.2 J / (g·ºC)

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  1. 20 June, 08:12
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    b) C = 0.50 J / (g°C)

    Explanation:

    Q = mCΔT

    ∴ Q = 50 J

    ∴ m = 10.0 g

    ∴ ΔT = 35 - 25 = 10 °C

    specific heat (C):

    ⇒ C = Q / mΔT

    ⇒ C = 50 J / (10.0 g) (10 °C)

    ⇒ C = 0.50 J / (g°C)
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