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Anonymous asked in Science & MathematicsChemistry · 2 months ago

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?

1 Answer

  • 2 months ago
    Favorite Answer

    Energy needed

    = m Cp ΔT

    = (52.0 g) × (4.184 J/g°C) × [(100.0 - 0.0)°C]

    = 21800 J

    = 21.8 kJ

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