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tino4ka555 [31]
3 years ago
15

When an aluminum bar is connected between a hot reservoir at 860 K and a cold reservoir at 348 K, 2.40 kJ of energy is transferr

ed by heat from the hot reservoir to the cold reservoir
(a) In this irreversible process, calculate the change in entropy of the hot reservoir.
_______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?
Physics
1 answer:
Natalija [7]3 years ago
6 0

Answer:

a) \Delta S_{in} = 2.791\,\frac{J}{K}, b) \Delta S_{out} = 6.897\,\frac{J}{K}, c) S_{gen} = 4.106\,\frac{J}{K}, d) Due to irreversibilities due to temperature differences.

Explanation:

a) The change in entropy of the hot reservoir is:

\Delta S_{in} = \frac{2400\,J}{860\,K}

\Delta S_{in} = 2.791\,\frac{J}{K}

b) The change in entropy of the cold reservoir is:

\Delta S_{out} = \frac{2400\,J}{348\,K}

\Delta S_{out} = 6.897\,\frac{J}{K}

c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:

\Delta S_{in} - \Delta S_{out} + S_{gen} = 0

S_{gen} = \Delta S_{out} - \Delta S_{in}

S_{gen} = 6.897\,\frac{J}{K} - 2.791\,\frac{J}{K}

S_{gen} = 4.106\,\frac{J}{K}

d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.

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A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p
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Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

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Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

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      u (1 + u ’v / c²) = v - u’

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Let's calculate

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We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

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Dennis_Churaev [7]

At the end of one full time period, the ant has returned to where it was at the beginning of the time period. Its displacement is <em>zero</em>.

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