Specific Heat of a metal.?

hi guys, would anybody help me to get the answer.

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  • 3 years ago
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    Energy transferred to water

    Q = mcDeltaT

    Q = 100g x 4..18 J g^-1K^-1 x(100 - 10)K

    Q = 100g x 4.18 J g^-1K^-1 x 90K

    Q = 150480J (transferred to the water from the metal).

    Energy transferred from the metal

    Q = mcDeltaT

    c = Q / mDeltaT

    c = 150480 / 10g x ( 250 - 20)K

    c = 150480 / (10g x 230K )

    c = 150480 J / 2300 gK

    c = 65.426 J g^-1 K^-1

    NB

    I have calculated using the modern S.I. nomenclature of units. ; Joules(J) & Kelvin(K)

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