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How much heat is needed to vaporize 33.3 grams of ethyl alcohol at its boiling point of 78.0°C? The latent heat of vaporization of ethyl alcohol is 857 J/g. Round your answer to three significant figures. joules

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The amount of heat needed to vaporize 33.3 grams of ethyl alcohol at its boiling point of 78.0°C is 28,500 joules.

Answer:

Q = 28538.1 J

Explanation:

As we know that heat require to vaporize the ethyl alcohol is given as

Q = mL

here we know that

m = mass = 33.3 gram

L = latent heat of fusion = 857 J/g

now from above formula

[tex]Q = (33.3 gram) \times (857 J/gram)[/tex]

[tex]Q = 28538.1 J[/tex]

so heat required in three significant figures is given as

[tex]Q = 2.85\times 10^4 J[/tex]

so heat required to vaporize ethyl alcohol is [tex]2.85\times 10^4 J[/tex]

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