Physics, 28.02.2020 00:53 madisyn13

A heat engine operates between a hih temp reservoir at 610K and low temp reservoir at 320K. In one cycle, the engine absorbs 6400J from the high temp reservoir and does 2200 J of work. What is the net change in entropy as a result of this cycle?

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To solve this problem it is necessary to apply the concepts related to entropy. Entropy can be defined as the change between heat energy and body temperature. For this case we will analyze the entropy in the cold body and later in the hot body. From there we will find the entropy difference

Entropy can be defined as,

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

Here,

Q = Heat Flow

T = Temperature

The entropy of the hot reservoir goes down by

$S_1 = \frac{6400J}{610K}$

The entropy of the "cool" reservoir increases by

$S_2 = \frac{4200J}{320K}$

$\Delta S = S_2 -S_1$

$\Delta S = \frac{4200J}{320K}-\frac{6400J}{610K}$

$\Delta S = 13.125-10.492$

$\Delta S = 2.633J/K$

As expected, entropy increased in this process; the entropy lost by the hot side was less than the entropy gained by the cool side.

Answer from: franvelpaulino2191

the net change in entropy as a result of this cycle is 2.63 J/K

Explanation:

given information:

high temperature, $T_{H}$ = 610 K

low temperature, $T_{C}$ = 320 K

heat absorbed, $Q_{H}$ = 6400 J

work, W = 2200 J

to calculate the net change in entropy as a result of this cycle, we can use the following formula:

ΔS = $\frac{Q_{H} }{T_{H} } +\frac{Q_{C} }{T_{C} }$ and W = $Q_{H}$ - $Q_{C}$

where

ΔS = the change of entropy

$T_{H}$ = the high temperature

$T_{C}$ = low temperature

$Q_{H}$ = absorbed heat

$Q_{C}$ = released heat

first, we determine the $Q_{C}$

W = $Q_{H}$ - $Q_{C}$

$Q_{C}$ = 6400 - 2200

= 4200 J

therefore,

ΔS = $\frac{Q_{H} }{T_{H} } +\frac{Q_{C} }{T_{C} }$

= $-\frac{6400}{610} + \frac{4200}{320}$

= - 10.49 + 13.125

= 2.63 J/K

Answer from: Quest

Ibelieve your answer is b.hope this : )

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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