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    can anyone help me with this question.

    q) The fusion of deuterium nuclei can be represented by

    2/1 H + 2/1 H = 4/2 He (top ones the mass and bottom is the atomic number)

    a) calculate energy released by this reaction in joules given that:
    mass of 2/1 H = 2.01419 u
    mass of 4/2 He = 4.00277
    1 u is equivalent to 1.49 x 10 ^-10 J

    b) 2kg (1000 mol) of deuterium are caused to fuse and it is proposed that the energy released by fusion is used to generate electricity in a power station.
    If the efficiency of the process were 52% and the electrical output of the station is to be 5.0 MW, how long would the deuterium last for?

    thanks
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    (Original post by vp03)
    can anyone help me with this question.

    q) The fusion of deuterium nuclei can be represented by

    2/1 H + 2/1 H = 4/2 He (top ones the mass and bottom is the atomic number)

    a) calculate energy released by this reaction in joules given that:
    mass of 2/1 H = 2.01419 u
    mass of 4/2 He = 4.00277
    1 u is equivalent to 1.49 x 10 ^-10 J

    b) 2kg (1000 mol) of deuterium are caused to fuse and it is proposed that the energy released by fusion is used to generate electricity in a power station.
    If the efficiency of the process were 52% and the electrical output of the station is to be 5.0 MW, how long would the deuterium last for?

    thanks
    a) E = (2*2.01419 - 4.00277)u = 3.82x10^-12 J

    b) 1000mol deuterium = 500mol He formed

    E/mol= E * 6.02x10^23 = 2.3x10^12 J mol^-1

    = 1.15x10^15 J for 2kg deuterium at 100% efficieny
    = 6.0x10^14 J at 52% efficiency

    t=E/P = 1.2x10^8 s


    Is this right?
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    yeh dats rite. but can u pls explain to me how u got that.
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    why did you go from 2.3 to 1.15 in b)??
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    a)

    Because of E=mc^2, the mass of each particle can be given in terms of energy (ie - in terms of 'u'). When the two deuterium nuclei collide to create an alpha particle (4/2 He), the mass of the He is slightly less than the combined masses of the original nuclei. This mass is lost as it is released as energy. To calculate the size of this energy, you just find the difference between the energies (mass equivalent) of the product and reactants. ie - E = (2*2.01419u - 4.00277u) = 3.82x10^-12 J

    b) This energy calculated above is for the formation of one helium nucleus. If 1000mol of deuterium nuclei are used, then 500mol of helium nuclei are created (equation: 2D --> 1He). Therefore, to calculate the energy released by this reaction, you multiply the energy of formation of one nucleus of He (from part a) by the number of nuclei formed (500moles= 500 * 6.02x10^23). This gives a value of E = 1.15x10^15 J. However, only 52% of this is converted to electrical energy so (0.52*1.15x10^15 = ) 6.0x10^14J of electrical energy is obtained.

    If the power output is 5x10^6W and power is energy per unit time, then the time this energy will last = E/P = 1.2x10^8s.
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    tnx a lot bruv. really gr8ful for dat.

    can u explain one last thing. y do u consider helium? i was doin it in terms of deuterium. tnx
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    The value calculated in part a) is the energy released when one helium nucleus is formed from two deuterium nuclei (energy of original deuterium nuclei - energy of helium nucleus formed).

    So to calculate the total energy, you multiply a) (the energy when one helium nucleus is formed from two deuterium nuclei) by the number of helium nuclei formed.
 
 
 
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