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4 years ago

6

A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a hot reservoir at 600 K and rejec

ts energy by heat transfer to a cold reservoir at temperature TC. Determine the energy rejected, in kJ, and TC, in K.

1 answer:

Arada [10]4 years ago

5 0

Answer:

Explanation:

thermal efficiency = 40%

efficiency = (T₂ -T₁ ) / T₂ where T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.

(600 -T₁ ) / 600 = .4

600 -T₁ = 240

T₁ = 360K

Energy converted into work = 50 x .4 = 20 kJ

heat rejected = 50 - 20 = 30 kJ

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3 years ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

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3 years ago

Strands of materials A and B are placed under a tensile force of 10 Newtons. Material A deforms more than Material B.

nadezda [96]

Answer:

True

Explanation:

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3 years ago

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The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi

Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,

1 gallon = 0.133 cubic feet (cf)

Therefore,

Capacity of Carvins Cove water reservoir in cf = 3.2 x 10˄9 x 0.133

= 4.28 x 10˄8

Applying Mass balance i.e

Accumulation = Mass In - Mass out (Eq. 01)

Here

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation = 0.5 - 11

= - 10.5 cfs

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

= 37,800

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

= 907,200

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days = (4.28 x 10˄8) x 1/907,200

= 471 Days.

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

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3 years ago

A distribution center is used in which of the following applications?

FrozenT [24]

both b and c are the right

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3 years ago