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1 June, 02:10

A 30.0 g sample of a metal was heated in a hot water bath to 80°c. it was then quickly transferred to a coffee-cup calorimeter. the calorimeter contained 100.0 g of water at a temperature of 20°c. the final temperature of the contents of the calorimeter was 25°c. what is the specific heat capacity of the metal?

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  1. 1 June, 04:05
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    when the metal is lost heat and the calorimeter of water is gained the heat

    and when the heat lost = the heat gained so,

    (M*C*ΔT) m = (M*C*ΔT) w

    when Mm = mass of the metal = 30 g

    Δ Tm = (80-25) = 55 °C

    and Mw = mass of water = 100 g

    Cw is the specific heat of water = 4.181 J/g.°C

    ΔTw = (25-20) = 5 °C

    so by substitution:

    ∴ 30 * Cm*55 = 100 * 4.181 * 5

    ∴Cm (specific heat of metal) = (100*4.181*5) / (30*55)

    ∴C of metal = 1.267 J/g.°C
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