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How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the tempature from 30 degrees celcius to 65 degrees celcius?

Respuesta :

Answer:

Q = 73.173 KJ

Explanation:

  • Q = m×C×ΔT

∴ C: water heat capacity

∴ C H2O (25°C) = 4.1813 J/g.K

∴ m H2O = 0.5 Kg = 500 g

∴ T1 = 30°C ≅ 303 K

∴ T2 = 65°C ≅ 338 K

⇒ ΔT = 338 - 303 = 35 K

⇒ Q = ( 500 g )×( 4.1813 J/g.K )×( 35 K )

⇒ Q = 73172.75 J = 73.173 KJ

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