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2 years ago

Nuclear reactors use fuel rods to heat water and generate steam. Is this process endothermic or exothermic?

Physics

2 answers:

lana66690 [7]2 years ago

7 0

Answer:

The heat produced by the nuclear reactor is an exothermic process while the heat absorbed by the water to convert into steam is an endothermic process.

Explanation:

Nuclear reactor being the heat of a nuclear power plant uses the radioactive uranium fuel to generate the heat by the process of nuclear fission in a controlled manner.

The processing of uranium is carried out into small ceramic pellets which are stacked together into sealed metal tubes known as fuel rods.

Usually more than 200 such rods are bunched together leading to the formation of a fuel assembly.

The core of the reactor is often made up of a couple hundred assemblies, according to its power level.

Inside the reactor vessel, these fuel rods are immersed into water which serve as both a coolant and moderator. The moderator helps slow down the neutrons produced by fission to sustain the chain reaction.

Anna71 [15]2 years ago

5 0

Explanation:

Exothermic reaction is defined as the reaction in which release of heat takes place. This also means that in an exothermic reaction, bond energies of reactants is less than the bond energies of products.

Hence, difference between the energies between the reactants and products releases as heat and therefore, enthalpy of the system will decrease.

Whereas in an endothermic reaction, heat is supplied from outside and absorbed by the reactant molecules. Hence, enthalpy of the system increases.

As water acts as a coolent and when fuel rods in a nuclear reactor are immersed in it then heat created by coolent is absorbed by water and then it changes into steam.

Since, absorption of heat occurs in the nuclear reactor. Therefore, it is an endothermic reaction.

Thus, we can conclude that nuclear reactors use fuel rods to heat water and generate steam. This process is endothermic.

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What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates

34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance $C_{0}$ = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed $3.00 \times 10^{4}$ V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

V = Ed

Now,

Q = CV

or, Q = $C \times Ed$

Therefore, substitute the values into the above formula as follows.

Q = $C \times Ed$

= $8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})$

= $2.55 \times 10^{-10} C$

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is $2.55 \times 10^{-10} C$ .

3 0

3 years ago

In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed

Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is $87.87 * 10^{-37}$ .

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= $5.29 * 10^{-11} m\\$

velocity= $2.9* 10^{6} m/s$

Working formula, M=N/A

$I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}$

$T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }$

= $15.16 * 10^{-5} s$

$I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }$ $= 0.10 * 10^{-14}$

= $1 * 10^{-15} C$

M= $1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}$

= $87.87 * 10^{-37}$

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

#SPJ4

4 0

9 months ago

Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma

Sloan [31]

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance ( $C_{HP}$ ) is given by

$C_{HP} = \dfrac{Q_{H}}{W_{in}}$

where, $Q_{H}$ is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature $T_{H}$ and $W_{in}$ is the required input and is given by $W_{in} = Q_{H} - Q_{L}$ , $Q_{L}$ being magnitude of heat transfer between cyclic device and low-temperature $T_{L}$ . Therefore, from above equation we can write,