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andrew-mc [135]

4 years ago

14

A heat engine is operating on a Carnot cycle and has a thermal efficiency of 75 percent. The waste heat from this engine is reje

cted to a nearby lake at 15 ⁰C at a rate of 14 kW. Determine the power output of the engine and the temperature of the heat source.

1 answer:

AysviL [449]4 years ago

8 0

Answer:

Power Output = 42 KW

T₁ = 1152 K = 879°C

Explanation:

The thermal efficiency of a Carnot's Engine is given by:

η = 1 - Q₂/Q₁

where,

η = efficiency = 75% = 0.75

Q₂ = Heat Rejection Rate = 14 KW

Q₁ = Heat Absorption Rate = ?

Therefore,

0.75 = 1 - 14 KW/Q₁

14 KW = (1 - 0.75)(Q₁)

Q₁ = 14 KW/0.25

Q₁ = 56 KW

Thus,

Power Output = Q₁ - Q₂

Power Output = 56 kW - 14 KW

<u>Power Output = 42 KW</u>

<u></u>

The thermal efficiency of a Carnot's Engine is also given by:

η = 1 - T₂/T₁

where,

η = efficiency = 75% = 0.75

T₂ = Temperature of Heat Sink (Lake) = 15°C + 273 = 288 K

T₁ = Temperature of Heat Sink = ?

Therefore,

0.75 = 1 - 288 K/T₁

288 K = (1 - 0.75)(T₁)

T₁ = 288 K/0.25

<u>T₁ = 1152 K = 879°C</u>

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Alona [7]

Answer:

C₂ = 2.22 KJ/kg °C

Explanation:

Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

$Heat\ Lost\ by\ Metal\ Block = Heat\ Gained\ by\ Ice\\m_{1}C_{1} \Delta T_{1} = m_{2}C_{2} \Delta T_{2}$

where,

m₁ = mass of ice = 1 kg

C₁ = specific heat of ice = 2.04 KJ/kg.°C

ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C

m₂ = mass of metal block = 1 kg

C₂ = specific heat of metal = ?

ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C

Therefore, using these values in the equation, we get:

$(1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_{2}(13.88\ ^0C) \\C_{2} = \frac{30.84\ KJ}{13.88\ kg/^0C}$

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4 0

3 years ago

Read 2 more answers

an oscilloscope has a time/div setting of 50ms and there are 10 divisions on the time scale. a sine wave on the oscilloscope dis

mamaluj [8]

Answer:

The frequency is f = 6Hz

Explanation:

Given

The time division setting is $d_t =50ms = 50*10^{-3}s$

The number of division is $n= 10$

The total time for 10 division can be mathematically obtained as

$T_{total} = 10 *50*10^{-3}$

$=500 *10^{-3}s$

From the question the time for 3 cycle is

$T_{total} = 500*10^{-3}s$

Then the for one cycle which equivalent to the Period $(T)$ $= \frac{T_{total}}{3}$

$= \frac{500 *10^{-3}}{3}$

The frequency is generally given as $f = \frac{1}{period} = \frac{1}{T}$

Now substituting values we have

$f = \frac{3}{500 *10^{-3}}$

$= 6Hz$

4 0

4 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (6)

Where:

$a=-4.9 m/s^{2}$

$b=48.1 m/s$

$c=1.5 m$

Substituting the known values:

$t=\frac{-48.1 \pm \sqrt{48.1^{2}-4(-4.9)(1.5)}}{2(-4.9)}$ (7)

Solving (7) we find the positive result is:

$t=9.847 s$ (8)

Substituting this value in (1):

$x=(48.1 m/s)cos(0\°) (9.847 s)$ (9)

$x=473.640 m$ This is the horizontal distance the cannonball traveled before it landed on the ground.

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3 years ago

Evaluate the expression a+b when a=23 and b=45 . Write in simplest form

Viefleur [7K]

If a = 23 and b = 45

a + b = ?

23 + 45 = 68

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3 years ago

If matter cannot be created nor destroyed the how was matter created

Assoli18 [71]

It can’t be created nor destroyed although it can be changed from one form to the other

I hope this helps

Have a happy holidays:)

7 0

3 years ago