A 0.867 kg sample of ethanol is in its liquid state at its boiling temperature of 78 C. If 1.93e5 J of energy are transferred into the sample, what percentage of the sample is vaporized?ethanol has:a specific heat of 1000 J/kgK in solid form,a specific heat of 2010 J/kgK in liquid form,a freezing temperature of –114 Ca boiling temperature of 78 C,a latent heat of fusion of 1.05e5 J/kg, anda latent heat of vaporization of 8.54e5 J/kg

Question
Asked Nov 15, 2019
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A 0.867 kg sample of ethanol is in its liquid state at its boiling temperature of 78 C. If 1.93e5 J of energy are transferred into the sample, what percentage of the sample is vaporized?
ethanol has:
a specific heat of 1000 J/kgK in solid form,
a specific heat of 2010 J/kgK in liquid form,
a freezing temperature of –114 C
a boiling temperature of 78 C,
a latent heat of fusion of 1.05e5 J/kg, and
a latent heat of vaporization of 8.54e5 J/kg
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Expert Answer

Step 1

Consider an ethanol of mass m having latent heat of vaporization Lv and Qgiven is the amount of heat is given to the sample.

m 0.867 kg
L-8.54x10 kg
Qgven =1.93x10 J
given
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m 0.867 kg L-8.54x10 kg Qgven =1.93x10 J given

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Step 2

Write the formula for heat(Qrequired) required to change its state to vapour from liquid of the entire ethanol and plug the required values to find it.

Qrequired mL
=(0.867 kg)(8.54x10 Jkg)
= 7.4x10 J
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Qrequired mL =(0.867 kg)(8.54x10 Jkg) = 7.4x10 J

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Step 3

Since, the heat given(Qgiven) to the substance is less than the heat required(Qrequired), therefore, only certain amount of substance change its state fr...

Qgven = m'L
given
т —
LL
(1.93x10 J
8.54x10 Jkg
т %3
m 0.226 kg
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Qgven = m'L given т — LL (1.93x10 J 8.54x10 Jkg т %3 m 0.226 kg

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