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ohaa [14]
3 years ago
14

SOMEONE HELP PLEASE 20 POINTS

Physics
2 answers:
professor190 [17]3 years ago
5 0
#3.
The answer would be C.

#4.
The answer would be B.

Sorry I didn't answer earlier! Just noticed that it just got answered. ):
VashaNatasha [74]3 years ago
3 0
Question three is C and question 4 is b

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What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i
babymother [125]

Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

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T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

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Substituting the values in the equation (1) we get

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or

\frac{Q_2}{3.78\times 10^3}=0.72

or

Q_2=3.78\times 10^3\times0.72

⇒ Q_2 =2.721\times 10^3 J

Now,

The entropy change (\Delta S) is given as:

\Delta S=\frac{\Delta Q}{T_1}

or

\Delta S=\frac{Q_1-Q_2}{T_1}

substituting the values in the above equation we get

\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}

\Delta S=1.69J/K

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