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Alika [10]
3 years ago
10

How much heat is required to vaporize 1.5kg of liquid water at its boiling point

Physics
1 answer:
Dafna11 [192]3 years ago
3 0

Given:

Water, 1.5kg at boiling point

Required:

Heat required to vaporize at boiling point

Solution:

Water at boiling point is at 100degrees Celsius

Changing it into Kelvin, just dd 273 to 100 = 373 K

H = mCpT

H = (1.5kg)(4.184kJ/kg-K)(373K)

<span>H = 2340.9kJ</span>

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Un vagón de 250 kg situado en la cima de una montaña rusa inicia su descenso por una rampa inclinado 60º sobre la horizontal. Si
earnstyle [38]
Um I don’t understand what your saying
6 0
2 years ago
If you calculate the thermal power radiated by typical objects at room temperature, you will find surprisingly large values, sev
OverLord2011 [107]

Answer:

best explanation of this is sentence B

Explanation:

The radiation emission of the bodies is given by the expression

     P = σ A e T⁴

Where P is the power emitted in watts, σ is the Stefan-Boltzmann constant, A is the surface area of ​​the body, e is the emissivity for black body e = 1 and T is the absolute body temperature in degrees Kelvin.

When the values ​​are substituted the power is quite high 2.5 KW, but the medium surrounding the box also emits radiation

   T box ≈ T room

    P box ≈ P room

As the two powers are similar and the box can absorbed, since it has the ability to emit and absorb radiation, as the medium is also close of the temperature of the box, the amount emitted is very similar to that absorbed, so the net change in energy is very small.

   In the case that the box is much hotter or colder than the surrounding medium if there is a significant net transfer.

Consequently, the best explanation of this is sentence B

5 0
3 years ago
A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi
kati45 [8]

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well  nitrate per 15 \mathrm{m}^{3} of water

nitrate in distant reservoir =5 \mathrm{~g} / \mathrm{m}^{3}

Let the flow rate of well

flow rate of reservoir =y m^{3} / s

Drinking water requirement is 45 \mathrm{ppm} or 45 \mathrm{~g} / \mathrm{m}^{3}

therefore, the total flow of drinking water

 

8 0
2 years ago
A solar panel measures 0.57 m by 1.3 m. In direct sunlight, the panel delivers 6.6 A at 6 V. If the intensity of sunlight is 300
zheka24 [161]

Answer:

efficiency of solar panel is 18%

Explanation:

Energy received by the sunlight is given as

Power = intensity \times area

here we know the dimensions of the plate as

L = 0.57 m

W = 1.3 m

now we have

Area = (0.57)(1.3) m^2

A = 0.741 m^2

now the power received by the sun light is given as

P_1 = 0.741(300) = 222.3 W

now the output power due to solar panel is given as

P_{out} = V i

P_{out} = (6.6 A)(6 V)

P_{out} = 39.6 Watt

now the efficiency is given as

\eta = \frac{P_{out}}{P_{in}}

\eta = \frac{39.6}{222.3}

\eta = 0.18

8 0
3 years ago
A 900 kg car moves around a 500 m radius curve at 25.0 m/s. What is the centripetal force on
Gnesinka [82]

Answer:

9375 N

Explanation:

From the question,

Centripetal force (F) = mv²/r.................. Equation 1

Where m = mass of the car, v = velocity of the car, r = radius of the curve.

Given: m = 900 kg, r = 600 m, v = 25 m/s

Substitute these values into equation 1

F = (900×25²)/600

F = 9375 N.

Hence the centripetal force on the car is 9375 N

3 0
3 years ago
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