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IrinaVladis [17]

1 year 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

1 answer:

asambeis [7]1 year 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

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See Explanation

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The question is incomplete; as the mixtures are not given.

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Where

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T = Amount of all elements/components

Take for instance:

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How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

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We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

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The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.

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