If 3.40 x 10^5 J of energy is supplied to a container of liquid oxygen at -183 ° C , how much oxygen can evaporate?

This is Giancoli Answers with Mr. Dychko. The amount of heat needed to change phase of a certain mass of the substance is the mass times the latent heat of vaporization in this case because it's going from a liquid to a gas. So, we'll divide both sides by L to solve for m. So, the amount of mass that can turn into gas is going to be the energy absorbed which is 3.4 times 10 to the 5 joules divided by the latent heat of vaporization for oxygen which is 210 times 10 to the 3 joules per kilogram which is about 1.6 kilograms.

COMMENTS

By soja78866 on Mon, 3/11/2019 - 3:34 PM

How did you get 2.1x10^5 j/kg? That number is not in the problem.

By Mr. Dychko on Tue, 4/2/2019 - 7:30 PM

This number is found in a data table 14-3 titled "Latent Heats" on page 397. It's the latent heat of vaporization for oxygen.

Find us on:

Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.

If 3.40×105 J3.40 \times 10^5 \textrm{ J}3.40×105 J of energy is supplied to a container of liquid oxygen at −183∘C-183 ^\circ \textrm{C}−183∘C, how much oxygen can evaporate?

Find us on:

If $3.40 \times 10^5 \textrm{ J}$ of energy is supplied to a container of liquid oxygen at $-183 ^\circ \textrm{C}$ , how much oxygen can evaporate?