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BabaBlast [244]
3 years ago
12

A wave is moving toward the shore If its frequency is 2.7 hertz, what is its wavelength? What would the correct number of signif

icant figures be?
Chemistry
1 answer:
TiliK225 [7]3 years ago
7 0
When multiplying and deviding follow the least number of sf.
since wavelength = 1/period
hence wavelength=1/2.7
=0.37 (2sf)
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A Carnot cycle operates between the temperatures limits of 400 K and 1600 K, and produces 3600 kW of net power. The rate of entr
TiliK225 [7]

The rate of entropy change:

The rate of entropy change of the working fluid during the heat addition process is 3 kW/K

What is the Carnot cycle?

  • The Carnot Cycle is a thermodynamic cycle made up of reversible isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression processes in succession.
  • The ratio of the heat absorbed to the temperature at which the heat was absorbed determines the change in entropy.

The entropy of a system:

The rate of heat addition is expressed as,

Q = \frac{WT_{H}}{T_{H}- T_{L}}

The entropy of a system is a measure of how disorderly a system is getting. The rate of entropy generation during heat addition is,

S_{gen} = \frac{Q}{T_{H}} = \frac{W}{T_{H} - T_{L}}

Calculation:

<u>Given:</u>

T_{L} = 400K

T_{H} = 1600K

W = 3600 kW

Put all the values in the above equation, and we get,

S_{gen} = \frac{W}{T_{H} - T_{L}} = \frac{3600}{1600-400} = 3 kW/K

The rate of entropy change is 3 kW/K

Learn more about the Carnot cycle here,

brainly.com/question/13002075

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2 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

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