A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 2.85×10−2 Kg. Throughout this problem use Lf=3.34*10^5 j/kg for the heat of fusion for water. - During this time, how much work is performed by the engine? my attempt: i noticed that the two temperature differences are 100 degrees and 0 degrees(mixture of icenwater) so the maximum efficiency is n = 1- qc/qh = 0.27 how much work is performed by the engine, Q(work)=Mass melted * latent heat fusion thing Qc= m*Lf so (2.85*10^-2 ) * (3.34*10^5) = 9519 Joules now this is where my question comes in, do i multiply the efficiency, 0.27 by 9519 to get the work performed by the engine or is that the work lost by the engine so 9519-.27*9519 i just wanted to check before i submit the awnser(it's mastering physics)
Go back to the definition of efficiency: W/Qh. In any engine, W = Qh-Qc. [tex]\eta = 1 - \frac{Q_c}{Q_h}[/tex] You have Qc. What you want to find first is Qh. AM
I'm sorry, i had to teach myself everything until i got to university, can you please explain how you solved for Qh, from n = 1-(qc/qh) so that i might not make that mistake again? and from this, Qh = (Qh - Qc) / n aren't i supposed to know qh to solve for qh? or is that equation the same as n=1-(qh/qc)
I used the solver in my calculator to find Qh from n = 1 - (qc/qh) and it was 13039J w = qh-qc, 13039 - 9519 = 3507 J which was correct but would still like to know how to rearrange, n = 1 - (qc/qh) to solve for qh =] thanks everyone
Your best friend is some pen and paper man... Flip some numbers around and do some mathematical acrobatics to achieve [tex]Q_h = \frac{Q_c}{1-\eta}[/tex]
Vorcil. You need to study algebra before you can hope to do anything here. Get a tutor - fast. It is not difficult but you need this basic skill. When you have an equation, you can rearrange it by doing the same thing to each side: [tex]\eta = 1 - \frac{Q_c}{Q_h}[/tex] subtract 1 from both sides and multiply both sides by -1: [tex]\frac{Q_c}{Q_h} = \eta - 1[/tex] Multiply both sides by Qh and divide by [itex]\eta - 1[/itex] [tex]\frac{Q_c}{\eta -1} = Q_h[/tex] AM
thanks for showing me, had no idea that you could solve it like that, as for studying calculus and algebra, i have reasonable skills, i'm doing first year math and physics papers and don't find anything hard except for the occasional wierd algebraic things that come up, i've mastered a majority of the basic algebraic skills needed to solve problems in physics but there are some rules that i've never seen before, for the example above, i knew you could subtract 1 from both sides, but didn't know you could use the Qh, multiply both sides by it to move it from one side of the equation to the other. after finals, i'll go on youtube, and watch all the special algebra movies on Mathtv, the guy explains most of the rules i missed out on learning cheers AM and that other fella