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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Answers

Answer 1

Answer:

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_(1) = u_(g)$ = 2553.6 kJ/kg

$v_(1) = v_(g) = 0.4625 m^(3)/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_(2) = v_(g) = 0.4625 m^(3)/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_(2) = v_(g) = 0.4625 m^(3)/kg$ and temperature $T_(2) = 360^(o)C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_(2) = u_{\text{at 5 bar, 400^(o)C}} + (\frac{v_(2) - v_{\text{at 5 bar, 400^(o)C}}}{v_{\text{at 7 bar, 400^(o)C - v_(at 5 bar, 400^(o)C)}}})(u_{at 7 bar, 400^(o)C - u_(at 5 bar, 400^(o)C)})$

$u_(2) = 2963.2 + ((0.4625 - 0.6173)/(0.4397 - 0.6173))(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$(Q)/(m) = \Delta u$

$(Q)/(m) = u_(2) - u_(1)$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

Answer 2

Answer:

Final answer:

The heat transfer is 227.4 kJ per kg of water.

Explanation:

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. To determine the heat transfer in kJ per kg of water, we need to calculate the heat absorbed by the water as it reaches 360°C.

Using the specific heat capacity of water (4,186 J/kg°C) and the change in temperature (360°C - 100°C), we can calculate the heat transfer:

Qw = mw * cw * AT = (1 kg) * (4186 J/kg°C) * (360°C - 100°C) = 227,440 J = 227.4 kJ

Therefore, the heat transfer is 227.4 kJ per kg of water.

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a difference in temperature. This fundamental phenomenon occurs through three main mechanisms: conduction, convection, and radiation. Conduction involves the direct transfer of heat through a material, such as metal. Convection is the transfer of heat through the movement of fluids (liquids or gases). Radiation is the emission of electromagnetic waves carrying heat energy. Understanding heat transfer is essential in various fields, including physics, engineering, and environmental science, as it governs temperature regulation, climate dynamics, and the functioning of countless technological devices.

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Related Questions

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe? 2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.a. the same as before. b. lower than before.c. higher than before.3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.A. Only the odd multiples of the fundamental frequency.B. Only the even multiples of the fundamental frequency.C. All integer multiples of the fundamental frequency.E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?A. Half the length of the open-open pipe.B. Twice the length of the open-open pipe.C. One-fourth the length of the open-open pipe.D. Four times the length of the open-open pipe.E. The same as the length of the open-open pipe.F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the thirdharmonic.
In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after are given. The collision is (Show your work-no work shown = ZERO POINTS) 1.8 m/s 0.2 m/s 0.6 m/s 1.4 m/s 4 kg 6 kg 4 kg 6 kg Before After A) perfectly elastic. B) partially inelastic. C) completely inelastic. D) characterized by an increase in kinetic energy E) not possible because momentum is not conserved.
(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?
Find the current if 20C of charge pass a particular point in a circuit in 10 seconds.
An object of mass m swings in a horizontal circle on a string of length L that tilts downward at angle θ.Find an expression for the angular velocity ω in terms of g, L and angle θ.

When a 5.0 kg box is hung from a spring, the spring stretches to 50 mm beyond its relaxed length. (a) In an elevator accelerating upward at 2.0 m/s2 , how far does the spring stretch with the same box attached? (b) How fast and in which direction should the elevator accelerate for the spring stretch to be zero (that is, the spring returns to its relaxed length)?

Answers

The extension of the spring in the elevator is 60 mm.

For the extension of the spring to be zero, the elevator must be moving downwards under free fall.

The given parameters;

mass of the box, m = 5 kg
extension of the spring, x = 50 mm = 0.05 m

The spring constant is calculated as follows;

F = kx

mg = kx

$k = (mg)/(x) \n\nk = (5 * 9.8)/(0.05) \n\nk = 980 \ N/m$

The tension on the spring in an elevator accelerating upwards is calculated as follows;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 2)

T = 59 N

The extension of the spring is calculated as follows;

$T = kx\n\nx = (T)/(k) \n\nx = (59)/(980) \n\nx = 0.06 \ m\n\nx = 60 \ mm$

For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.

For free fall, a = g

T = m(g - a) = 0

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Answer:

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

we know that;

mg = kX

5 × 9.81 = k(0.05)

k = 981 N/m

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

A convex lens is placed on a flat glass plate and illuminated from above with monochromatic red light. When viewed from above, concentric bands of red and dark are observed. What does one observe at the exact center of the lens where the lens and the glass plate are in direct contact?A) a darkspotB) a bright spot thatis some color other than redC) a bright redspotD) a rainbow of color

Answers

Answer:

After passing through the glass plate, the red light disperses and meets at point.

The convex lens has two refracting surfaces, and convex kens is called as converging lens. So, at the exact center of the lens, one observes a Dark spot.

Thus, the correct option is a) one observes a dark spot.

Answer:

The answer is: A) a darkspot

Explanation:

When the red light passes through the glass plate, it is scattered. the convex lens (convergent lens) has two refractive surfaces, therefore, in the center of the lens, a characteristic dark spot would be observed.

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) $v=5.6725\,m.s^(-1)$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^(-2)$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^(-1)$

putting the values in eq. (1)

$v^2=0^2+2* 3.62 * 10$

$v=8.5088\,m.s^(-1)$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20* 8.5088$

$p=170.1764\,kg.m.s^(-1)$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)* v'=p$

$(20+10)* v'=170.1764$

$v'=5.6725\,m.s^(-1)$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$(1)/(2) (M+m).v'^2=(M+m).g.h$

$(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h$

$h\approx 1.6420\,m$

above the normal hanging position.

A spring has a spring constant of 1350 N/m. You place the spring vertically with one end on the floor. You then drop a 1.3 kg book onto it from a height of 0.8 m above the top of the spring. Find the maximum distance the spring will be compressed. Express your answer with the appropriate mks units.

Answers

Answer:

0.123 m.

Explanation:

From Hook's law,

The potential energy of the book = the energy stored in the spring.

mgh = 1/2ke².................. Equation 1

Where m= mass of the book, g = acceleration due to gravity, h = height, k = spring constant of the spring, e = distance of compression.

make e the subject of the equation

e = √(2mgh/k).................. Equation 2

Given: m = 1.3 kg, h = 0.8 m, k = 1350 N/m

Constant: g = 9.8 m/s²

Substitute into equation 2

e = √(2×1.3×0.8×9.8/1350)

e = √(20.384/1350)

e = √(0.0151)

e = 0.123 m.

Answer:

0.015m (downwards)

Explanation:

When the book is dropped on the top of the spring at that height, the potential energy ( $E_(P)$ ) of the book is converted to elastic energy ( $E_(E)$ ) on the spring thereby causing a compression on the spring. i.e

$E_(P)$ = $E_(E)$

But;

The potential energy $E_(P)$ of the mass (book), is the product of the mass(m) of the book, the height(h) from which it was dropped and the acceleration due to gravity (g). i.e

$E_(P)$ = - m x g x h [the -ve sign shows a decrease in height as the mass (book) drops]

Also;

The elastic energy ( $E_(E)$ ) of compression of the spring is given by

$E_(E)$ = $(1)/(2)$ x k x c

Where;

c = compression length of the spring

k = the spring's constant

Substitute these values of $E_(P)$ and E into equation (i) as follows;

- m x g x h = $(1)/(2)$ x k x c ----------------(ii)

From the question;

m = 1.3kg

h = 0.8m

Take g = 10m/s²

k = 1350N/m

Substitute these values into equation (ii) as follows;

- 1.3 x 10 x 0.8 = $(1)/(2)$ x 1350 x c

- 10.4 = 675c

Solve for c;

c = - 0.015 m [The negative sign shows that the spring actually compresses]

Therefore, the maximum distance the spring will be compressed is 0.015m (downwards of course).

Two point charges have a total electric potential energy of -24 J, and are separated by 29 cm.If the total charge of the two charges is 45 μC, what is the charge, in μC, on the positive one?
What is the charge, in μC, on the negative one?

Answers

Answer:

The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Explanation:

Electric potential energy between two point charges is derived from concept of Work, Work-Energy Theorem and Coulomb's Law and described by the following formula:

$U_(e) = (k\cdot q_(A)\cdot q_(B))/(r)$ (1)

Where:

$U_(e)$ - Electric potential energy, measured in joules.

$q_(A)$ , $q_(B)$ - Electric charges, measured in coulombs.

$r$ - Distance between charges, measured in meters.

$k$ - Coulomb's constant, measured in kilogram-cubic meters per square second-square coulomb.

If we know that $U_(e) = -24\,J$ , $q_(A) = 45* 10^(-6)\,C+ q_(B)$ , $k = 9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2))$ and $r = 0.29\,m$ , then the electric charge is:

$-24\,J = -(\left(9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2)) \right)\cdot (45* 10^(-6)\,C+q_(B))\cdot q_(B))/(0.29\,m)$

$-6.96 = -405000\cdot q_(B)-9* 10^(9)\cdot q_(B)^(2)$

$9* 10^(9)\cdot q_(B)^(2)+405000\cdot q_(B) -6.96 = 0$ (2)

Roots of the polynomial are found by Quadratic Formula:

$q_(B,1) = 1.327* 10^(-5)\,C$ , $q_(B,2) \approx -5.827* 10^(-5)\,C$

Only the first roots offer a solution that is physically reasonable. The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

A car traveling at 45 km/h starts to brake, and comes to a stop over a distance of 18 m. Calculate the accelerationof the braking car.