Change the chapter
On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at $0^\circ\textrm{C}$ and completely melts to $0^\circ\textrm{C}$ water in exactly one day? 1 watt = 1 joule / second (1W = 1 J/s).
Question by OpenStax is licensed under CC BY 4.0.

$13.5 \textrm{ W}$

Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 14, Problem 15 (Problems & Exercises) (0:55)

Sign up to view this solution video!


1 vote with a rating of 5

Quiz Mode

Why is this button here? Quiz Mode is a chance to try solving the problem first on your own before viewing the solution. One of the following will probably happen:

  1. You get the answer. Congratulations! It feels good! There might still be more to learn, and you might enjoy comparing your problem solving approach to the best practices demonstrated in the solution video.
  2. You don't get the answer. This is OK! In fact it's awesome, despite the difficult feelings you might have about it. When you don't get the answer, your mind is ready for learning. Think about how much you really want the solution! Your mind will gobble it up when it sees it. Attempting the problem is like trying to assemble the pieces of a puzzle. If you don't get the answer, the gaps in the puzzle are questions that are ready and searching to be filled. This is an active process, where your mind is turned on - learning will happen!
If you wish to show the answer immediately without having to click "Reveal Answer", you may . Quiz Mode is disabled by default, but you can check the Enable Quiz Mode checkbox when editing your profile to re-enable it any time you want. College Physics Answers cares a lot about academic integrity. Quiz Mode is encouragement to use the solutions in a way that is most beneficial for your learning.

Calculator Screenshots

OpenStax College Physics, Chapter 14, Problem 15 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. A cooler contains 3.50 kilograms of ice which has a temperature of zero degree Celsius and in exactly one day which we converted into seconds by multiplying by 24 hours per day and then 3600 seconds per hour. Ammm….. in that time we have all this ice melting and to a question is what is the rate of vartanity entering ice or the power transfer into the ice. So, that’s going to be the amount of energy required to melt that much ice divided by the time in which its melting. So, the amount of energy to melt ice will be the mass of ice multiply by the latent heat of fusion for water and then…... so, we have 3.5 kilograms times 334 times ten to three joules per kilogram divided by 86400 seconds. In this case 13.5 watts.