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Harrizon [31]
3 years ago
10

Determine whether the sign of heat or work (relative to the system) is positive or negative. A metal cylinder is rolled up a ram

p. (The metal cylinder is the system.) The fireman extinguishes the fire with foam and cools the environment. (The foam is the system.)
Chemistry
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

Metal cylinder: Work is <em>positive</em>

Foam: Heat is <em>positive</em>.

Explanation:

Work is done as a resultant force acts along a displacement. Heat is the flow of energy from one body to another, resulting in a either a rise or a drop in temperature of the bodies. Work and Heat are forms of energy and they can either be assigned a positive or negative sign with respect to a thermodynamic system.

If work is done on a body (thermodynamic system), the work is negative. On the other hand, if work is done by the body, it is termed as a positive work. Heat is positive when it flows into a thermodynamic system and negative when it is lost by the system.

In this case, though the metal cylinder is rolled up a ramp, the metal cylinder does a positive work along the displacement. Considering the foam, heat is gained by the foam (thermodynamic system) as it cools the environment, thus, heat is positive.

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(The radioisotope 224Ra decays by alpha emission via two paths to the ground state of its daughter 94% probability of alpha deca
pav-90 [236]

Answer:

a) ²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b) the Q-value of this reaction is 5.789 MeV

c) the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

       

Explanation:

a)

The decay equation for the alpha decay is expressed as;

²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b)

Calculate the Q-value (in MeV) of this reaction.

Q = Mparent - Mdaughter -Mg

Q = MRa - MRn -Mg

= 224.020202 - 220.011384 - 4.00260305

= 0.00621495 amu

= 5.789 MeV

therefore the Q-value of this reaction is 5.789 MeV

c)

Energy of alpha particle is expressed as;

E∝ = MQ / ( m + M)

now this is the maximum energy available for the daughter, ²²⁰Rn going to the ground state;

The energy of the alpha particle gives;

E∝  = 220(5.789) / ( 4 + 220) = 5.69 MeV

as given in the question,The other less frequent alpha occurring 5.5% of the time leaves the daughter nucleus in an excited state of 0.241 MeV above the ground state.

Therefore the energy of this alpha is

E∝ = 5.69 - 0.241 = 5.449 Mev

Therefore the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

d)

Sketch of the nuclear decay scheme have been uploaded along side this answer.

7 0
3 years ago
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