How much heat must be removed from 75.0 g of water at 90.0°C to cool it to 23.0°C?

Eyeslandic asked Apr 21, 2023

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Eyeslandic asked Apr 21, 2023

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10 votes

Answer:

See below

Step-by-step explanation:

75 g * ( 90 - 23) C * 1 cal / g-C

= 5025 cal = 21 034 .7 J

Tim Gradwell answered Apr 27, 2023

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