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.

Amount of heat = mass of water * specific heat (temperature change

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= 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

When molecules of the reactant hit each other, only a certain percentage of the collisions cause a chemical change.If the molecules are moving too slowly, they will bounce off each other. There will be no reaction.If the molecules are moving rapidly but do not collide with the correct orientation, they will again bounce off each other. There will be no reaction.

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I hope this helps 

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Zane