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1 May, 16:11

A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water and calorimeter were initially at 21.0oC. If the final temperature of mixture was 30.5oC, calculate the specific heat capacity of the metal. The specific heat of water is 4.184 J / (g. oC) and heat capacity of calorimeter is 10.0 J/oC.

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  1. 1 May, 16:23
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    Cp = 0.44 J/g. C

    This is heat capacity of metal.

    Explanation:

    From energy conservation

    Heat lost by metal = Heat gain by water + Heat gain by calorimeter

    Because here temperature of metal is high that is why it loose the heat. The temperature of water and calorimeter is low that is why they gain the heat.

    final temperature is T = 30.5 C

    We know that sensible heat transfer given as

    Q = m Cp ΔT

    m=Mass

    Cp=Specific heat capacity

    ΔT=Temperature difference

    By putting the values

    55 x Cp (99.5 - 30.5) = 40 x 4.184 (30.5 - 21) + 10 x (30.5 - 21)

    Cp (99.5 - 30.5) = 30.65

    Cp = 0.44 J/g. C

    This is heat capacity of metal.
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