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Explain the terms nuclear fission and specific heat capacity - Leaving Cert Physics - Question 8 - 2018

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Explain the terms nuclear fission and specific heat capacity. Water can act as both a moderator and a coolant in a nuclear fission reactor. What effect does a mode... show full transcript

Worked Solution & Example Answer:Explain the terms nuclear fission and specific heat capacity - Leaving Cert Physics - Question 8 - 2018

Step 1

Explain the terms nuclear fission and specific heat capacity.

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Answer

Nuclear fission is the process of splitting a large nucleus into two smaller nuclei, accompanied by the emission of energy and neutrons. It releases a significant amount of energy, which is harnessed in nuclear reactors.

Specific heat capacity is the amount of energy required to change the temperature of 1 kg of a substance by 1 K. It indicates how a substance can store thermal energy.

Step 2

What effect does a moderator have on the rate of fission?

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Answer

A moderator increases the rate of fission by slowing down the neutrons produced during the fission reaction. This increases the likelihood of these slower neutrons being absorbed by other fissile nuclei, promoting more fission reactions.

Step 3

How does a moderator have this effect?

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A moderator works by reducing the kinetic energy of fast neutrons. This is achieved through collisions between the neutrons and the atoms of the moderator material, effectively slowing them down to thermal velocities where they can efficiently sustain the fission chain reaction.

Step 4

Calculate the energy absorbed by the water.

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Answer

The energy absorbed by the water can be calculated using the formula:

E=mimescimeshetaE = m imes c imes heta

Where:

  • m=5000extkgm = 5000 ext{ kg}
  • c=2.23imes106extJ/kgc = 2.23 imes 10^6 ext{ J/kg} (specific latent heat of vaporisation)
  • heta=70extK heta = 70 ext{ K} (temperature increase)

Substituting in these values:

E=(5000extkg)imes(4180extJ/kgK)imes(70extK)E = (5000 ext{ kg}) imes (4180 ext{ J/kg K}) imes (70 ext{ K})

Calculating:
E=1.26imes1010extJE = 1.26 imes 10^{10} ext{ J}

Step 5

Write a nuclear equation for this reaction.

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Answer

92235U+n56139Ba+3694Kr+3n^{235}_{92}U + n \rightarrow ^{139}_{56}Ba + ^{94}_{36}Kr + 3n

Step 6

Calculate the energy released, in MeV, in this reaction.

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Answer

To calculate the energy released during fission, we can use the mass-energy equivalence formula:

E=mc2E = mc^2

Where the mass loss in this reaction is given as 3.0 x 10^{-28} kg. Substituting in the value for c (the speed of light, approximately 3.00imes108extm/s3.00 imes 10^8 ext{ m/s}):

E=(3.0imes1028extkg)×(3.00imes108extm/s)2E = (3.0 imes 10^{-28} ext{ kg}) \times (3.00 imes 10^8 ext{ m/s})^2

After calculations, we find:

E=2.74imes1011extJE = 2.74 imes 10^{-11} ext{ J}

Converting this to MeV (1 J = 6.242imes10126.242 imes 10^{12} MeV), we get:

E=171extMeVE = 171 ext{ MeV}

Step 7

Explain why fusion reactors are not yet a practical source of energy on Earth.

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Answer

Fusion reactors are not currently a practical source of energy because the conditions required for fusion to occur—extremely high temperatures and pressures—are difficult to achieve and maintain. Moreover, the energy input needed to create and sustain these conditions often exceeds the energy produced.

Step 8

Give one other advantage that a fusion reactor would have over a fission reactor.

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Answer

One major advantage of fusion reactors over fission reactors is that they produce significantly less radioactive waste, making them more environmentally friendly. Additionally, the fuel sources for fusion (like isotopes of hydrogen) are more abundant than those for fission.

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