Ask question

Ask question

All categories

3 years ago

11

106 grams of liquid water are in a cylinder with a piston maintaining 1 atm (101325 Pa) of pressure. It is exactly at the boilin

g point of water, 373.15 K. We then add heat to boil the water, converting it all to vapor. The molecular weight of water is 18 g/mol and the latent heat of vaporization is 2260 J/g. 1) How much heat is required to boil the water?

1 answer:

xeze [42]3 years ago

4 0

Answer:

239.55 KJ

Explanation:

Given:

Mass 'm' = 106 g

Latent heat of vaporization'L'= 2260 J/g.

Molecular weight of water'M' = 18 g/mol

Pressure 'P' = 101325 Pa

Temperature 'T' = 373.15 K

Using the formula of phase change, in order to determine the amount of heat required, we have

Q = mL

Q = 106 x 2260

Q = 239560J = 239.55 KJ

You might be interested in

Determine the power that needs to besupplied by the fanifthe desired velocity is 0.05 m3/s and the cross-sectional area is 20 cm

Mariulka [41]

Answer:

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

Explanation:

Complete statement is: <em>Determine the power that needs to besupplied by the fan if the desired velocity is 0.05 cubic meters per second and the cross-sectional area is 20 square centimeters.</em>

From Thermodynamics and Fluid Mechanics we know that fans are devices that work at steady state which accelerate gases (i.e. air) with no changes in pressure. In this case, mechanical rotation energy is transformed into kinetic energy. If we include losses due to mechanical friction, the Principle of Energy Conservation presents the following equation:

$\eta\cdot \dot W = \dot K$

$\dot W = \frac{\dot K}{\eta}$ (Eq. 1)

Where:

$\eta$ - Efficiency of fan, dimensionless.

$\dot W$ - Electric power supplied fan, measured in watts.

$\dot K$ - Rate of change of kinetic energy of air in time, measured in watts.

From definition of kinetic energy, the equation above is now expanded:

$\dot W = \frac{\rho_{a}\cdot \dot V}{2\cdot \eta}\cdot \left(\frac{\dot V}{A_{s}} \right)^{2}$ (Eq. 2)

Where:

$\rho_{a}$ - Density of air, measured in kilograms per cubic meter.

$\dot V$ - Volume flow, measured in cubic meters per second.

$A_{s}$ - Cross-sectional area of fan, measured in square meters.

If we know that $\rho_{a} = 1.20\,\frac{kg}{m^{3}}$ , $\dot V = 0.05\,\frac{m^{3}}{s}$ , $\eta = 0.3$ and $A_{s} = 20\times 10^{-4}\,m^{2}$ , the power needed to be supplied by the fan is:

$\dot K = \left[\frac{\left(1.20\,\frac{kg}{m^{3}} \right)\cdot \left(0.05\,\frac{m^{3}}{s} \right)}{2\cdot (0.3)} \right]\cdot \left(\frac{0.05\,\frac{m^{3}}{s} }{20\times 10^{-4}\,m^{2}} \right)^{2}$

$\dot K = 62.5\,W$

A fan with an energy efficiency of 30 % would need 62.5 watts to bring a desired volume flow of 0.05 cubic meters per second through a cross-sectional area of 20 square centimeters.

5 0

3 years ago

A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of

mina [271]

Answer:

(orbital speed of the satellite) V₀ = 3.818 km

Time (t) = 4.5 × 10⁴s

Explanation:

Given that:

The radius of the Earth is 6.37 × 10⁶ m; &

the acceleration of gravity at the satellite’s altitude is 0.532655 m/s

We can calculate the orbital speed of the satellite by using the formula:

Orbital Speed (V₀) = √(r × g)

radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m

= (2.1 × 10⁷ + 6.37 × 10⁶) m

= 27370000

= 2.737 × 10⁷m

Orbital Speed (V₀) = √(r × g)

Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )

= 3818.215

= 3.818 × 10³

= 3.818 Km

To find the time it takes to complete one orbit around the Earth; we use the formula:

Time (t) = 2 π × $\frac{r}{V_o}$

= 2 × 3.14 × $\frac{2.737*10^7}{3.818*10^3}$

= 45019.28

= 4.5 × 10 ⁴ s

6 0

3 years ago

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

$\frac{13}{20z+r} =7.6$ .......(1)

$\frac{13}{40z+r} =4.5$ ......(2)

Substituting the value of r from (2) in (1), we have,

$13=152z+7.6\times\frac{13-180z}{4.5}$

which simplifying gives us, $z=0.0589\Omega/m\approx0.06\Omega/m$ (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 $\Omega$

3 0

2 years ago

What is the average electric current in a wire, when a charge of 150 C passes in 30 s?

NARA [144]

We know, I = Q / t
Here, Q = 150 C
t = 30 s

Substitute their values,
I = 150 / 30
I = 5 A

In short, Your Answer would be 5 Ampere

Hope this helps!

7 0

3 years ago

Hello....I need help with this question.

musickatia [10]

Answer:

3.2 m/s²

Explanation:

Acceleration can be calculated as:

v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time)

25 m/s = 9 m/s + a(5 s) (a is unknown)

16 m/s = a(5 s)

a = 3.2 m/s²

We assume that this is a uniform acceleration (meaning that the velocity increases at an equal rate for those 5 seconds).

6 0

3 years ago

Read 2 more answers