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7 April, 10:31

suppose that during the icy hot lab that 65 kj of energy were transferred to 450 g of water at 20 C. What would have been the final temperature of the water

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  1. 7 April, 10:54
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    The final temperature of water is 54.5 °C.

    Explanation:

    Given dа ta:

    Energy transferred = 65 Kj

    Mass of water = 450 g

    Initial temperature = T1 = 20 °C

    Final temperature = T2 = ?

    Solution:

    First of all we will convert the heat in Kj to joule.

    1 Kj = 1000 j

    65 * 1000 = 65000 j

    specific heat of water is 4.186 J / g. °C

    Formula:

    q = m * c * ΔT

    ΔT = T2 - T1

    Now we will put the values in Formula.

    65000 j = 450 g * 4.186 J / g. °C * (T2 - 20°C)

    65000 j = 1883.7 j / °C * (T2 - 20°C)

    65000 j / 1883.7 j / °C = T2 - 20°C

    34.51 °C = T2 - 20°C

    34.51 °C + 20 °C = T2

    T2 = 54.5 °C
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