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
r is 4.18 J/gºC.
2 answers:
=4.18*80*(75-45)
=4.18*80*30
=10032
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
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