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3 years ago

15

A Carnot engine has a power output of 148 kW. The engine operates between two reservoirs at 17◦C and 727◦C. How much thermal ene

rgy is absorbed each hour? Answer in units of J.

1 answer:

Elanso [62]3 years ago

3 0

Answer:

The thermal Energy absorbed each hour = 750420 kJ

Explanation:

Efficiency of a Carnot engine = W/Q = (Th-Tc)/Th ............... Equation 1.

W/Q = (Th-Tc)/Th

making Q the subject of the equation above,

Q = W(Th)/(Th-Tc)........................ Equation 2

Where Q = input power, W = output power, Th = temperature of the hot reservoir, Tc = Temperature of the cold reservoir.

<em>Given: W= 148 kW, Th = 727 °C = 727 + 273 = 1000 K, Tc = 17 °C =17 + 273 = 290 K.</em>

<em>Substituting these values into equation 2,</em>

<em>Q = 148(1000)/(1000-290)</em>

<em>Q = 148000/710</em>

Q = 208.45 kw

The thermal Energy absorbed each hour = 208.45 kW × 3600 s

The thermal Energy absorbed each hour = 750420 kJ

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4vir4ik [10]

Answer: $a=2.42 m/s^2$

Explanation:

Given

mass $m_1=3.50\times 10^{3} kg$

$m_2=1.00\times 10^{3} kg$

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$\mu =0.21$

Let T be the tension in the rope

From Diagram

$m_1g\sin \theta -T-f_r=m_1a$ -----------------1

where $f_r=friction\ force$

$f_r=\mu m_g\cos \theta$

For block $m_2$

$T=m_2a$ -----------2

From 1 & 2

$m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a$

$m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a$

$\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a$

$a=2.42 m/s^2$

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4 years ago

(a) In the baggage claim area of an airport, a particular baggage carousel is shaped like a section of a large cone, steadily ro

Tom [10]

Answer:

Part a)

$F_f = 107.8 N$

Part b)

$\mu = 0.415$

Explanation:

Part a)

Time period of one revolution is given as

T = 42 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{42}$

$\omega = 0.15 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.15^2)(7.46)cos20.5 = F_f$

$F_f = 107.8 N$

Part b)

New Time period of one revolution is given as

T = 30 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{30}$

$\omega = 0.21 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.21^2)(7.94)cos20.5 = F_f$

$F_f = 112.9 N$

Also we know that in perpendicular direction also force is balanced

$F_n + m\omega^2 r sin\theta = mgcos\theta$

$F_n = mgcos\theta - m\omega^2 r sin\theta$

$F_n = 30(9.81)cos20.5 - 30(0.21)^2(7.94)sin20.5$

$F_n = 272 N$

now for friction coefficient we will have

$F_f = \mu F_n$

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3 years ago

Calculate the average acceleration of a car that changes speed from 0 m/s to 20 m/s in 5 s.

Vikki [24]

Answer:

4

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2 years ago

In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its

Mila [183]

Answer: hello below is the missing part of your question

A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.

answer

x = 0.0962 m

Explanation:

<em>First step :</em>

Determine the length of the rough patch/spot

F = Uₓ (mg)

and w = F.d = Uₓ (mg) * d

hence;

d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m

<em>next : </em>

work done on unstretched spring length

Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m

w' = Uₓ (mg) * d

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also given that the Elastic energy of spring = work done ( w')

1/2 * kx^2 = 23.27 J

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4 years ago