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.

Question

Mattie Bartlett

Chemistry

11 December, 06:50

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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Answers (1)

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Fernanda Fuller · Answer 1

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