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The Cp of liquid water is 4.184 J/g°C. How much energy is needed to heat a 52.0 gram sample of water from 0.0°C to 100.0°C?
- Uncle MichaelLv 72 months agoFavorite Answer
= m Cp ΔT
= (52.0 g) × (4.184 J/g°C) × [(100.0 - 0.0)°C]
= 21800 J
= 21.8 kJ