The familiar "ether" used as an anesthetic agent is diethly ether...

Question

The familiar "ether" used as an anesthetic agent is diethly ether, C4H10O. Its heat of vaporization is +26.5 kJ/mol at Calorimetry and Heat Capacity (Section 9.7)
9.76 What is the difference between heat capacity and specific heat? 9.77 Does a measurement carried out in a bomb calorimeter give
its boiling point. How much energy in kilojoules is required to convert 100 mL of diethyl ether at its boiling point from liquid to vapor if its density is 0.7138 g/mL?

Accepted Answer

Hi everyone. This problem reads the term alcohol generally refers to the compound ethanol or ethyl alcohol. The heat of vaporization for ethanol is 38.6 kg joules per mole if the combustion of ethanol is carried out in an a diabolic calorie emitter. Can its boiling point be measured? How much energy and kill a jewel units is required to evaporate 15.4 g of ethanol at its boiling point from the liquid state to the gaseous state. The density of liquid ethanol is 0.0.790 grams per milliliter. So for this problem, we have two things we want to answer. The first is, can its boiling point be measured? And the second thing we want to answer is how much energy and kill a jewel units is required to evaporate 15.4 g of ethanol. Okay, so let's go ahead and talk about an a diabetic calorie meter. Okay, so an a diabolic calorie emitter, an example of one of these calorie emitters is the coffee cup calorie emitter. Okay, so and a diabetic process is one where no exchange of heat or mass takes place between the system and the surroundings. Okay, so an example is the coffee cup calorie emitter and in this type of calorie emitter there is no exchange of heat or mass. Okay, so a coffee cup calorie emitter works under constant pressure Under these conditions, the coffee cup calorie meter can be used to measure change and entropy. Okay, so change in entropy but the boiling point cannot be determined using an a diabetic calorie meter. Okay, so for this problem Can its boiling point be measured? No boiling point cannot be measured. So that is going to be the answer for part one, Part two asks us how much energy and kill a jewel units is required. Okay, so we're gonna start off with what we're given. Okay, so we're given 4.4 15. g of ethanol. So that's going to be our starting point. So we have 15.4 g of ethanol. And we want to go from massive ethanol to heat required to evaporate 15.4 g of ethanol. Alright, so what we're going to do is we first want to go from grams of ethanol, two moles of ethanol. So that means we're going to need the molar mass. Okay, so in one mole of ethanol and the ethanol, the uh, the formula for ethanol is C two H five oh H. And one mole of ethanol using the periodic table. The mass is 46.74 g of ethanol. Making sure our units cancel properly, are grams of ethanol cancel. And now we have molds of ethanol. We want to go from molds of ethanol to heat required. So what we're going to do is we're gonna go back to our problem and look at the value for the heat of vaporization for ethanol. And in the problem we're told the heat of vaporization for ethanol is 38.6 kg joules per mole. So we're gonna go ahead and plug that in here. So there is 38. kill a jewels of heat per one mole of ethanol. Okay. And for the beginning I wrote out ethanol and then I've also used the symbol, the formula. So those it's going to be the same thing. Okay, so our molds of ethanol cancel and now we're left with killer joules of heat. So let's go ahead and do this calculation. And when we do we get 12.9 kg jewels. Okay, so our final answer for this problem is going to be heat required equals 12.9 kg. So we just answered both parts of this question. The first part being, can its boiling point be measured? No. And the second part, How much energy and kill a jewel units is required? And that is 12.9 kg jewels. That is it for this problem? I hope this was helpful

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