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29 March, 03:57

Water (2190 g) is heated until it just begins to boil. If the water absorbs 5.75*105 J of heat in the process, what was the initial temperature of the water?

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  1. 29 March, 07:55
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    37.18°C.

    Explanation:

    The amount of heat absorbed by water = Q = m. c.ΔT.

    where, Q is amount of heat absorbed by water (Q = 5.75 x 10⁵ J).

    m is the mass of water (m = 2190 g).

    c is the specific heat capacity of liquid water = 4.18 J/g°C.

    ΔT is the temperature difference = (final T - initial T = 100.0°C - initial T).

    ∴ Q = m. c.ΔT = 5.75 x 10⁵ J.

    5.75 x 10⁵ J = (2190 g) (4.18 J/g°C) (100.0°C - initial T)

    ∴ (100.0°C - initial T) = (5.75 x 10⁵ J) / (2190 g) (4.18 J/g°C) = 62.81°C.

    ∴ initial T = (100.0°C - 62.81°C) = 37.18°C.
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