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2 years ago

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(b) How much energy must be supplied to boil 2kg of water? providing that the specific latent heat of vaporization of water is 3

30 kJ/kg.

1 answer:

Lelu [443]2 years ago

8 0

Complete question:

(b) How much energy must be supplied to boil 2kg of water? providing that the specific latent heat of vaporization of water is 330 kJ/kg. The initial temperature of the water is 20 ⁰C

Answer:

The energy that must be supplied to boil the given mass of the water is 672,000 J

Explanation:

Given;

mass of water, m = 2 kg

heat of vaporization of water, L = 330 kJ/kg

initial temperature of water, t = 20 ⁰C

specific heat capacity of water, c = 4200 J/kg⁰C

Assuming no mass of the water is lost through vaporization, the energy needed to boil the given water is calculated as;

Q = mc(100 - 20)

Q = 2 x 4200 x (80)

Q = 672,000 J

Q = 672,000 J

Q = 672,000 J

Therefore, the energy that must be supplied to boil the given mass of the water is 672,000 J

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

3 years ago

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

7 0

3 years ago

A bag of cement has a mass of 62 g. What is the mass of the bag of cement in S.I. units (kg)?

NNADVOKAT [17]

The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.

<u>Given the following data:</u>

Mass of cement = 62 grams.

To calculate the mass of this bag of cement in S.I. units (kg):

<h3>How to convert to S.I. units.</h3>

In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:

<u>Conversion:</u>

1000 grams = 1 kilograms.

62 grams = X kilograms.

Cross-multiplying, we have:

X = $\frac{62}{1000}$

X = 0.062 kilograms.

Read more on mass here: brainly.com/question/13833323

8 0

2 years ago

A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm

____ [38]

Answer:

53.13 °

Explanation:

In order to do this, we just need to apply the following:

tanα = Dy/Dx

Where:

Vy: speed of the ball in the y axis.

Vx: speed of the ball in the x axis.

At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.

According to this, then:

tanα = (40/30)

tanα = 1.3333

α = tan⁻¹(1.3333)

<h2>α = 53.13°</h2>

This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.

Hope this helps

6 0

2 years ago

The maximum energy a bone can absorb without breaking is surprisingly small. For a healthy human of mass 60 kg60 kg, experimenta

netineya [11]

Answer:

<em>the maximum height a man can jump from and land rigidly upright on both feet without breaking his legs is 0.34 m</em>

<em></em>

Explanation:

Mass of a healthy man = 60 kg

energy the bone can take without breaking = 200 J

If a healthy man jumps from a height 'h', he falls with an energy equal to the potential energy due to his initial height above the ground.

initial potential energy of the healthy man = mgh

where m = mass of the man

g = acceleration due to gravity = 9.81 m/s^2

h = the height above ground

==> PE = 60 x 9.81 x h = 588.6h

If we assume that all energy is absorbed in the leg bones in a rigid landing, then we can safely say that this calculated PE for a healthy man is equal to the energy his bone can absorb in the jump without breaking.

equating, we have

200 = 588.6h

<em>the maximum height a man can jump from without breaking his legs = 200/588.6 = 0.34 m</em>

When people jump from a height, the sudden deceleration to zero can impact a big force on the leg bones, shattering them. If the time spent in decelerating to zero is increased, the overall force on the leg bones is reduced greatly.

<em>Bending the knees gradually on landing from a jump from a height, and then rolling increases the time spent decelerating, and reduces the impact force on the legs due to the landing</em>. If you observe carefully you will see that this is what professional stunts men and acrobats do when they jump from a height.

5 0

3 years ago