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trasher [3.6K]
2 years ago
6

(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.
Physics
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 rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
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\omega=0.31\frac{rad}{s}

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Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

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A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
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Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

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       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
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