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3 June, 16:44

Water (2410 gg) is heated until it just begins to boil. If the water absorbs 4.95*105 JJ of heat in the process, what was the initial temperature of the water?

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  1. 3 June, 17:03
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    51.1°C

    Explanation:

    Parameters given:

    Mass of water = 2410g = 2.41 kg

    Heat absorbed = 4.95 * 10^5 J

    The formula for heat capacity in terms of mass, specific heat capacity and temperature is given as:

    H = mC (T2 - T1)

    Where m = mass

    C = specific heat capacity of water = 4200 J/kg°C

    T2 = final temperature

    T1 = initial temperature

    In this case, the final temperature is 100°C since the water boils, hence, to find T1,

    4.95 * 10^5 = 2.41 * 4290 * (100 - T1)

    => 100 - T1 = 495000 / (2.41 * 4200)

    100 - T1 = 48.9

    => T1 = 100 - 48.9

    T1 = 51.1°C

    The initial temperature of the water was 51.1°C
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