54.56 g of water at 80.4 oC is added to a calorimeter that contains 47.24 g of water at 40 oC. If the final temperature of the system is 59.

Question

54.56 g of water at 80.4 oC is added to a calorimeter that contains 47.24 g of water at 40 oC. If the final temperature of the system is 59.4 oC, what is the calorimeter constant (C calorimeter)

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Diễm Thu 5 years 2021-07-18T17:40:52+00:00 1 Answers 23 views 0

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  1. Tryphena
    0
    2021-07-18T17:42:19+00:00
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    Answer:

    49.5J/°C

    Explanation:

    The hot water lost some energy that is gained for cold water and the calorimeter.

    The equation is:

    Q(Hot water) = Q(Cold water) + Q(Calorimeter)

    Where:

    Q(Hot water) = S*m*ΔT = 4.184J/g°C*54.56g*(80.4°C-59.4°C) = 4794J

    Q(Cold water) = S*m*ΔT = 4.184J/g°C*47.24g*(59.4°C-40°C) = 3834J

    That means the heat gained by the calorimeter is

    Q(Calorimeter) = 4794J – 3834J = 960J

    The calorimeter constant is the heat gained per °C. The change in temperature of the calorimeter is:

    59.4°C-40°C = 19.4°C

    And calorimeter constant is:

    960J/19.4°C =

    49.5J/°C

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