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QveST [7]
3 years ago
13

Consider that 168.0 J of work is done on a system and 305.6 J of heat is extracted from the system. In the sense of the first la

w of thermodynamics, What is the value (including algebraic sign) of W, the work done by the system?
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0

To solve this problem we must resort to the Work Theorem, internal energy and Heat transfer. Summarized in the first law of thermodynamics.

dQ = dU + dW

Where,

Q = Heat

U = Internal Energy

By reference system and nomenclature we know that the work done ON the system is taken negative and the heat extracted is also considered negative, therefore

W = -168J \righarrow  Work is done ON the system

Q = -305.6J \rightarrow Heat is extracted FROM the system

Therefore the value of the Work done on the system is -158.0J

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Answer:

The change on the second particle is 2.93\times 10^{-16}\ C.

Explanation:

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It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

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If both particles take the same amount of time to go once around their respective circles. So,

T_e=T_s\\\\\dfrac{2\pi m_e}{Bq_e}=\dfrac{2\pi m_s}{Bq_s}\\\\\dfrac{m_e}{q_e}=\dfrac{m_p}{q_s}\\\\q_s=\dfrac{m_pq_e}{m_e}\\\\q_s=\dfrac{1.67\times 10^{-27}\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}\\\\q_s=2.93\times 10^{-16}\ C

So, the change on the second particle is 2.93\times 10^{-16}\ C.

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