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31 January, 17:21

if 980 KJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be

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  1. 31 January, 17:29
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    The final temperature of the water will be 328.81 K.

    Explanation:

    Using the equation, q = mcΔT

    here, q = energy

    m = mass

    c = specific heat capacity

    ΔT = change in temperature

    Mass of water = 1kg (1000 g) per liter

    ∴ 6.2 Liter of water = 6200 g

    c of water ≈ 4.18 J / g/K

    Now,

    980000 = 6200*4.18*ΔT

    ΔT = 37.81 K

    ∴ final temperature of the water = 291 + 37.81 = 328.81 K
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