I've attached a J-peg with the question in, laid out much more clearly than I could expect to type it here. I hope the attachment is ok.
The Attempt at a Solution
So far I have:
(1) dS = dQ / T
Entropy is a measure of a system's disorder.
(2) I've drawn a flow diagram but can't get a picture of that on here. Suffice to say I have tried to do this part quite hard but can't get it to work.
(a) dS = dQ /T ... dQ = 3/2 N K dT ... so dS = 3/2 N K dT/T. Then I can work out the integral and do this first part for temperature going from 293 to 273? Does mass not matter here?
(b) dS = dQ/T ... We know Q from the question as 333kJ/kg so S = Q/T goes to S = 333 * 1 kilogram of water / 273 Kelvin. Is this right?
(c) I have no idea how to do this.
The entropy change of the universe will increase during this process, as it will do for all processes above 0 Kelvin.
(4) Coefficient of Performance = Qh / Work = Qh / Qh - Qc ... this can then be written as Th / Th - Tc = 323K / 323K - 273K = 6.46.
To improve the coefficient of performance of the ice-maker you could decrease the temperature of the hot reservoir???? This doesn't seem to make sense...
If anyone has the time to look at this and help me I would very much appreciate it as I have lost all my second year notes on thermodynamics and can't see how to do this question properly.
45.9 KB Views: 301