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

asked May 13, 2023 31.5k views

12 votes

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5 votes

Answer:

See below

Step-by-step explanation:

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

= 5025 cal = 21 034 .7 J

DWoldrich answered May 17, 2023

3.0k points

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