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The specific heat of water is 4,186 J/kg°C. Approximately how much heat must be removed from 0.500 kg of water to change its temperature from 24.0°C to 5.00°C?

Answer :

Explanation:

Q= mc∆T

∆T= 5-24=- 19

Q= 0.5*4186*-19

Q= -39767 J

negative sign show heat releases

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