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16 March, 15:33

Exactly 253.0 J will raise the temperature of 10.0 g of a metal from 25.0 °C to 60.0 °C. What is the specific heat capacity of the metal?

a. 0.723 J / (g. °C)

b. 1.38 J / (g°C)

C 12.2 J/g °C)

d. 60.5J/g - °C)

e. None of these

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  1. 16 March, 16:09
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    Answer:a. 0.723 J / (g. °C)

    Explanation:

    Q = mCΔT

    where,

    Q - Quantity of heat supplied- = 253.0j

    m - mass of substance=10g

    C - specific heat capacity=?

    ΔT - change in temperature = 60 - 25 = 35°C

    Imputing values, we have,

    Q = mCΔT

    C = Q / (mΔT) = 253.0 / (10 x 35) = 0.7228 rounded to - --> 0.723 J / (g. °C)
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