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21 April, 12:12

At 1 atm, how much energy is required to heat 75.0 g H 2 O (s) at - 20.0 ∘ C to H 2 O (g) at 119.0 ∘ C?

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  1. 21 April, 13:48
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    238,485 Joules

    Explanation:

    The amount of energy required is a summation of heat of fusion, capacity and vaporization.

    Q = mLf + mC∆T + mLv = m (Lf + C∆T + Lv)

    m (mass of water) = 75 g

    Lf (specific latent heat of fusion of water) = 336 J/g

    C (specific heat capacity of water) = 4.2 J/g°C

    ∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C

    Lv (specific latent heat of vaporization of water) = 2,260 J/g

    Q = 75 (336 + 4.2*139 + 2260) = 75 (336 + 583.8 + 2260) = 75 (3179.8) = 238,485 J
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