Ask question

Ask question

All categories

4 years ago

8

A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

erature of the hot reservoir remains the same, what must be the temperature of the cold reservoir in order to increase the efficiency to 35.0%?

1 answer:

VLD [36.1K]4 years ago

4 0

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

You might be interested in

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 38 m/s when i

inysia [295]

Explanation:

look !

speed= 38m/s

start from rest= 0

5 0

3 years ago

A cat rides a merry-go-round while turning with uniform circular motion. At time t1 = 2.00 s, the cat's velocity is v with arrow

goldenfox [79]

Answer:

Part a)

$a_c = 2.07 m/s^2$

Part b)

$a_{avg} = 1.32 m/s^2$

Explanation:

As we know that it makes half revolution in given time interval

so we have

$\frac{T}{2} = t_2 - t_1$

$\frac{T}{2} = 9 - 2$

$T = 14 s$

now the angular speed is given as

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

$\omega = \frac{2\pi}{14}$

$\omega = 0.448 rad/s$

now linear speed is given as

$v = \sqrt{2.30^2 + 4.00^2}$

$v = 4.61 m/s$

now we have

$v = R \omega$

$4.61 = R(0.448)$

$R = 10.3 m$

Now centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = 0.448^2 \times 10.3$

$a_c = 2.07 m/s^2$

Part b)

Average acceleration of the cat is given as

$a_{avg} = \frac{v_2 - v_1}{\Delta t}$

$a_{avg} = \frac{2v}{\Delta t}$

$a_{avg} = \frac{2(4.61)}{9 - 2}$

$a_{avg} = 1.32 m/s^2$

7 0

4 years ago

An ice skater has a moment of inertia of 2.5 kg.m2 when her arms are extended and a moment of inertia of 1.2 kg.m2 when her arms

Damm [24]

Answer:

$\omega=4.7916\ rad.s^{-1}$

Explanation:

Given:

moment of inertia of the skater with extended arms, $I'=2.5\ kg.m^2$

moment of inertia of the the skater with pulled-in arms, $I=1.2\ kg.m^2$

angular velocity of the skater with extended arms, $\omega'=2.3\ rad.s^{-1}$

<u>Using the law of conservation of angular momentum:</u>

$I'.\omega'=I.\omega$

$2.5\times 2.3=1.2\times \omega$

$\omega=4.7916\ rad.s^{-1}$

6 0

3 years ago

Jaime lifts a package weighing 75N. if she lifts it 1.2 m, what work has she done

9966 [12]

Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J

6 0

3 years ago

What does it mean to say mass and weight are preportional to each other?

11Alexandr11 [23.1K]

It means that if the mass of a mass is increased by some factor, and nothing else changes, then the weight of that mass increases by exactly the same factor.

4 0

4 years ago