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16 April, 14:25

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?

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  1. 16 April, 14:48
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    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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