Answer to Question #134376 in Mechanics | Relativity for Jessica

Question #134376

A mass m = 0.043 kg of benzene vapor (Lv = 3.94x105 J/kg) at its boiling point of 80.1°C is to be condensed by mixing with water at 32.0°C. What is the minimum mass of water required to condense all of the benzene vapor? Assume the mixing and condensation take place is a perfectly insulating container.


1
Expert's answer
2020-09-22T17:10:07-0400

Let us determine the minimal amount of energy needed to be taken away from the benzene to condensate it:

"Q_- = L_v \\cdot m = 3.94\\cdot10^5\\,\\mathrm{J\/kg}\\cdot 0.043\\,\\mathrm{kg} \\approx 1.7\\cdot10^4\\,\\mathrm{J}." If we take away this amount, all the benzene will condensate and it will have the same temperature of 80.1°C.

Now we'll determine the amount of water, that can heat from 32.0°C to 80.1°C due to obtained heat "Q_-" :

"m_w c_w \\Delta T \\Rightarrow m_w = \\dfrac{Q_-}{c_w\\Delta T} = \\dfrac{1.7\\cdot10^4\\,\\mathrm{J}}{4.2\\cdot10^3\\,\\mathrm{J\/kg\/K}\\cdot48.1\\,\\mathrm{K}} = 0.084\\,\\mathrm{kg}."

If we take such an amount of water, the water will heat up to 80.1°C and the benzene will condensate. So the temperature of mixture will be 80.1°C.

If we take smaller amount of water, it will heat up to 80.1°C, but not all the benzene will condensate.

If we take larger amount, he benzene will condensate and cool.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS