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

)

= 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

samuel de champlain. in 1608, king henry iv sponsored pierre dugua, sieur de mons and samuel de champlain as founders of the city of quebec with 28 men. this was the second permanent french settlement in the colony of canada.