What amount of heat is removed to lower the temperature of 80 grams of water from 75ºC to 45ºC? The specific heat of liquid wate

Question

3 years ago

12

r is 4.18 J/gºC.

2 answers:

chubhunter [2.5K] · Answer 1 · 2022-08-03T16:40:34+03:00

5 0

=4.18*80*(75-45)

=4.18*80*30

=10032

amm1812 · Answer 2 · 2022-08-03T15:13:28+03:00

amm18123 years ago

3 0

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