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2 years ago

12

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:

chubhunter [2.5K]2 years ago

5 0

=4.18*80*(75-45)

=4.18*80*30

=10032

amm18122 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

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