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IrinaVladis [17]
2 years ago
7

nitial mass of water, at its boiling point, is 0.8 kg. 4 kW of heater is used to boil it completely. Assuming the specific laten

t heat of vaporization of water is 2MJ/kg, what is the time taken to vaporize all the water
Chemistry
1 answer:
asambeis [7]2 years ago
6 0

The  time taken to vaporize all the water is 400 seconds.

<h3>Total heat required to vaporize the water</h3>

The total heat required to vaporize the water is calculated as follows;

E = Lm

where;

  • L is latent heat of vaporization of water
  • m is mass of water

E  = 2 x 10⁶ J/kg x 0.8 kg

E = 1,600,000 J

<h3>Time taken to vaporize all the water</h3>

The time taken to vaporize all the water is calculated as follows;

E = power x time

time = E/power

time = ( 1,600,000 J) / (4,000 J/s)

time = 400 seconds

Thus, the  time taken to vaporize all the water is 400 seconds.

Learn more about heat of vaporization of water here: brainly.com/question/26306578

#SPJ1

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How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ
Jet001 [13]

Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

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r = 7.12 mm

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