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01:49 University of New Mexico

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Problem 127

The heat of vaporization of water at 373 K is 40.7 kJ>mol. Find q, w, E, and H for the evaporation of 454 g of water at this temperature at 1 atm.

$$952 \mathrm{kJ}$$

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## Video Transcript

we have 454 grams off water on the heat off vaporization off water is 373 caldron. It's 40.7 calo Jules Mahmoud so the number off moles can be the dome in as weight divided by more low Mars. The Moon of Mars off water is 18 substituting in the values we have. 454 ground off water divided by the molar mass. Off 18.16 The number off mole turn out to be 25 point morning. We all know that the heat Q is equal to and number off moves multiplied with Delta H that this the heat off vaporization, substituting in the values into expression. The heat will be a boat. 1030 Calo Jules here Q is equal to Delta H, but we know that Q was recorded 1030 Calo jewels, which also makes Delta H as but 1030 calo Jules. The equation for vaporization off water is represented on def. One more lawful liquid water is taken, then one moral off water people will be produced when 25.2 moles off liquid is taken the same amount off paper will be produced leading us to the book Done as 78.2. Hello, Jules. The change in internal energy for this process can be shown by the formula Delta E Change in energy. He'd less work done. We know that the value of Q is equal to 10 30 kilo Jews on the work done this negative 78 point dorky locals, substituting in the values we have 1030 minus 78 point two. So the change in an orgy will be 952. Hello, Jules.