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5 May, 11:21

Calculate the amount of heat released when the bottle (250g) of water is cooled from 30°C to 25°C. The specific heat of water is 4.186J/g Degree Celsius

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  1. 5 May, 11:41
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    5232.5J

    Explanation:

    Data obtained from the question. This includes the following:

    Mass (M) = 250g

    Initial temperature (T1) = 30°C

    Final temperature (T2) = 25°C

    Change in temperature (ΔT) = T1 - T2 = 30°C - 25°C = 5°C

    Specific heat capacity (C) = 4.186J/g°C

    Heat (Q) =.?

    The heat released can be obtained as follow:

    Q = MCΔT

    Q = 250 x 4.186 x 5

    Q = 5232.5J

    Therefore, the heat released is 5232.5J
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