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13 April, 18:48

1. You have an unknown amount of water vapor floating in a sealed container at 126℃. After 30 min the container is cooled to - 26℃, and a solid sample of water is found at 2.5g. What amount of energy was used for this process to occur?

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  1. 13 April, 21:07
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    7.8 kJ. The process used 7.8 kJ.

    The container is sealed, so

    mass of water vapour = mass of water = mass of ice.

    ∴ Heat required = heat to cool vapour + heat to condense vapour to water + heat to cool water + heat to freeze water + heat to cool ice

    q = q_1 + q_2 + q_3 + q_4 + q_5

    Step 1. Calculate q_1

    q_1 = mC_1ΔT_1

    m = 2.5 g; C_1 = 1.996 J.°C^ (-1) g^ (-1)

    ΔT_1 = T_f - T_i = 100 °C - 126 °C = - 26 °C

    q_1 = 2.5 g * 1.996 J.°C^ (-1) g^ (-1) * (-26°C) = - 130 J

    The negative sign shows that heat is removed from the system.

    Step 2. Calculate q_2

    q_2 = mΔ_cH

    Δ_cH = - 2257 J·g^ (-1)

    q_2 = 2.5 g * [-2257 J·g^ (-1) ] = - 5640 J

    Step 3. Calculate q_3

    q_3 = mC_3ΔT_3

    C_3 = 4.182 J.°C^ (-1) g^ (-1); ΔT_3 = T_f - T_i = 0 °C - 100 °C = - 100 °C

    q_3 = 2.5 g * 4.182 J.°C^ (-1) g^ (-1) * (-100 °C) = - 1050 J

    Step 4. Calculate q_4

    q_4 = mΔ_cH

    Δ_cH = - 334 J·g^ (-1)

    q_4 = 2.5 g * [-334 J·g^ (-1) ] = - 835 J

    Step 5. Calculate q_5

    q_5 = mC_5ΔT_5

    C_5 = 2.090 J.°C^ (-1) g^ (-1); ΔT_3 = T_f - T_i = - 26 °C - 0 °C = - 26 °C

    q_5 = 2.5 g * 2.090 J.°C^ (-1) g^ (-1) * (-26 °C) = - 136 J

    Step 6. Calculate the total heat involved

    q = q_1 + q_2 + q_3 + q_4 + q_5

    = (-130 - 5640 - 1050 - 835 - 136) J = - 7800 J = - 7.8 kJ

    The process removed 7.8 kJ.
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