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27 January, 20:56

In a coffee-cup calorimeter, 50.0 g hot water at 60.0 C was mixed with 50.0 g cold water at 20.0 C. If the final temperature is 36.3, what is the heat capacity of the calorimeter in J/C? Specific heat of water is 4.184 J / (g C)

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  1. 28 January, 00:46
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    2.88 J/g°C

    Explanation:

    Mass calorimeter (Mc) = 50g

    Temperature of calorimeter (T1) = 60°C

    Mass of water (Mw) = 50g

    Specific heat capacity of water (Cw) = 4.184 J/g°C

    Specific heat capacity of calorimeter (Cc) = ?

    Temperature of water (T2) = 20°C

    Final temperature of the content (T3) = 36.6°C

    Assuming no heat lose from the set up

    Heat loss by copper calorimeter = heat gain by the water

    Mc * Cc * (T1 - T3) = Mw * Cw * (T3 - T2)

    50 * Cc * (60 - 36.3) = 50 * 4.184 * (36.3 - 20)

    50 * Cc * 23.7 = 209.2 * 16.3

    1185Cc = 3409.96

    Cc = 3409.96 / 1185

    Cc = 2.877 J/g°C

    Cc = 2.88 J/g°C

    The specific heat capacity of the copper calorimeter is 2.88 J/g°C
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