Sapling (week 5 and 6) #2

[Jessica Luong 3K]

"A heat source generates heat at a rate of 83.0 W (1 W=1 J/s) . How much entropy does this produce per hour in the surroundings at 27.6 ∘C ? Assume the heat transfer is reversible."

I have already converted the temperature and the watts to Joules per hour, but I am still getting the wrong answer. Am I missing a step?

[Jiwon_Chae_3L]

Assuming that you're using the equation deltaS = q(reversible)/T, if you converted all of your units to joules per hour and kelvin, you should be getting a correct answer I think. You might be making a calculation error somewhere. Personally, I often forget to fully convert seconds to minutes to hours, instead only multiplying the term by 60 and leaving it in minute form.

[Mohamed Mido]

Watt's units is J/s so did you make sure to convert seconds to hours?

It should be something like this:
83 W = 83J/s x 3600 s/ 1 hr x 1/(27.6+273)K = 994 J/K*hr

[Evie Li_1H]

You would need to convert (q J/s x 60s/1 min x 60 min/hour) all over change in temperature in Kelvin (just add 273.15 to the degrees in Celsius to find Kelvin.)

[Samudrala_Vaishnavi 3A]

You probably found the number of joules produced in a second rather than an hour. Try cross multiplying or setting up a ratio essentially saying if in 1 second, (blank/answer in seconds from calc.)J produced, then in 3600 seconds (blank) J produced.

[ellenulitsky Dis 1I]

You first have to convert J/s to J/Hr and then divide (J/Hr)/(Degree in C + 273K) to get the units J/K*Hr. Hope this helps!

[Daniela Santana 2L]

Hi! In order to solve this problem take the heat being generated and divide it by the temperature (in kelvin). Then multiply the answer you get by 60 and then multiply that answer by 60 again. You should get the correct answer to the problem. I hope this helped :)

[VeronicaShepherd3B]

Do we know why our temperature has to be converted tot kelvin?