a heat input of 800J for each 1000J of work output, design 2 a heat input of 2500J, and design 3 a heat input of 3500J. Which design would you choose and why?
2006-06-21 06:07:50 · 5 answers · asked by jlynn 1 in Science & Mathematics ➔ Physics
Design 1 because it requires the least energy meaning the least risk of getting electrocuted when using a high voltage engine.The users safety must be considered too.
2006-06-21 06:11:47 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Design 1 is impossible, input heat must be greater than work done.
Design 2 is most efficient because it requires 2500 J whish means 1500 J will be lost to the sink.
2006-06-21 06:13:53 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋
Your forgeting needed info like wat the 450k is (money, temp, area, etc) I would go with the one that uses the less Heat input thus less work/force/ whatever to operate.
2006-06-21 06:12:27 · answer #3 · answered by Hitzumaru 2 · 0⤊ 0⤋
800joule in and 1000joule out? Thermo says its impossible.
2006-06-21 06:15:22 · answer #4 · answered by michael k 1 · 0⤊ 0⤋
the first
It is more efficient thermally
2006-06-21 06:10:17 · answer #5 · answered by KaizerSose 3 · 0⤊ 0⤋