Question

What amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC? The specific heat of liquid water is 4.18 J/gºC.

Answer 1

=4.18*80*(75-45)

=4.18*80*30

=10032

Answer 2

amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC with specific heat of liquid water is 4.18 J/gºC

= 4.18 J/gºC * 80g * (75-45)ºC

= 10032J