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tatiyna
3 years ago
10

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

t to a 42.4 degC reservoir?
Physics
1 answer:
babymother [125]3 years ago
7 0

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:

η =0.280

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

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

0.28=1-\frac{Q_2}{3.78\times 10^{3}}

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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\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

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