Question

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.

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