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25 December, 14:46

How many calories must be absorbed by 20.0 g of water to increase its temperature from 383.0 C to 303.0 C

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  1. 25 December, 17:15
    0
    1600 calories

    Explanation:

    Data obtained from the question include:

    M (mass of water) = 20g

    T1 (initial temperature) = 383°C

    T2 (final temperature) = 303°C

    ΔT (change in temperature) = T1 - T2 = 383 - 303 = 80°C

    C (specific heat capacity of water) = 1 calorie/g°C

    Q (heat) = ?

    Using the the equation Q = MCΔT, the heat in calories absorbed by the water can be obtained as follow:

    Q = MCΔT

    Q = 20 x 1 x 80

    Q = 1600 calories

    Therefore, the heat absorbed by the water is 1600 calories
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