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GrogVix [38]
3 years ago
13

A piston in a heat engine does 500 joules of work, and 1400 joules of heat are added to the system. Determine the change in inte

rnal energy and explain how this example demonstrates the conservation of energy.
Chemistry
1 answer:
san4es73 [151]3 years ago
8 0

Answer : The change in internal energy is, 900 Joules.

Solution : Given,

Heat given to the system = +1400 J

Work done by the system = -500 J

Change in internal energy is equal to the sum of heat energy and work done.

Formula used :

\Delta U=q+w

where,

\Delta U = change in internal energy

q = heat energy

w = work done

As per question, heat is added to the system that means, q is positive and work done by the system that means, w is negative.

Now put all the given values in the above formula, we get

\Delta U=(+1400J)+(-500J)=900J

Therefore, the change in internal energy is 900 J.

The change in internal energy depends on the heat energy and work done. As we will change in the heat energy and work done, then changes  will occur in the internal energy. Hence, the energy is conserved.

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If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera
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Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

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What is the frequency of a radio wave with a a wavelength of 3 m? (Hint: MHz
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a. 1 x 10^8

Explanation:

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I need help with this plz help
schepotkina [342]

Answer:

\rm ^{103}_{\phantom{1}40}Zr, zirconium-103.

Explanation:

In a nuclear reaction, both the mass number and atomic number will conserve.

Let ^{A}_{Z}\mathrm{X} represent the unknown particle.

The mass number of a particle is the number on the upper-left corner. The atomic number of a particle is the number on its lower-left corner under the mass number. For example, for the particle ^{A}_{Z}\mathrm{X}, A is the mass number while Z while Z is the atomic number.

Sum of mass numbers on the left-hand side of the equation:

\underbrace{239}_{^{239}_{\phantom{2}94}\mathrm{Pu}} + \underbrace{1}_{^{1}_{0}\mathrm{n}} = 240.

Note that there are three neutrons on the right-hand side of the equation. Sum of mass numbers on the right-hand side:

\underbrace{A}_{^{A}_{Z}\mathrm{X}} + \underbrace{134}_{^{134}_{\phantom{2}54}\mathrm{Xe}} + \underbrace{3\times 1}_{3\;^{1}_{0}\mathrm{n}} = A + 137.

Mass number conserves. As a result,

A + 137 = 240.

Solve this equation for A:

A = 103.

Among the five choices, the only particle with a mass number of 103 is \rm ^{103}_{\phantom{1}40}Zr. Make sure that atomic number also conserves.

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3 years ago
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