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1 June, 06:54

How many kilojoules are required to heat 0.500 g of water from 20.3 degrees Celcius to 29.7 degrees Celcius? How many Calories? Show your work and watch sig figs.

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  1. 1 June, 08:22
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    Amount of heat = mass of water * specific heat (temperature change

    )

    = 0.500 g * 4.184 J / g-C (29.7 - 20.3) C

    = 19.6648 J

    = 0.0197 KJ

    And

    1 cal = 4.186798 J

    19.6648 J * 1 cal / 4.186798 J = 4.70 cal
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