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3 years ago

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Determine the heat energy required to vaporize 13.9 grams of liquid water at 100° C. O 2,006 cal O 47.8 cal O 7,506 cal O 24.9 c

al

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

7506 cal

Explanation:

Given: Mass of water = m = 13.9 grams.

Heat energy required to determine the conversion of 13.9 grams of water to steam or vapor is m L,

where L = Latent heat of vaporization of water=540 cal/g.

Q = mL = (13.9)(540) = 7,506 cal.

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3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

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Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

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Now,

Kinetic energy of an electron = thermal kinetic energy

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$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

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$v_{p}$ = velocity of an proton

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Kinetic energy of a proton = thermal kinetic energy

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$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

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