# Heat engine- calculating minimum energy per cycle from hot reservoir

1. Aug 8, 2012

### twinklestar28

1. The problem statement, all variables and given/known data

A heat engine operating between freezing and boling points of water takes 20 cycles to raise a mass of 1000kg through a height of 3m. Calculate the minimum energy per cycle that must be extracted from the hot reservoir in order to achieve this.

2. Relevant equations

η=1-Tc/Th
η=W/Qh
P.E=mgh

3. The attempt at a solution

I assumed that the acceleration woas 9.81m/s^2 so I got 2943J for the Work done and then divided by 20 to get the energy for one cycle to get 147.15J. Plugging into the equation i got Qh= 525.5J which doesn't seem to sound right considering the question is worth 4 marks.Can i assume that the GPE of the mass is the work done or is there more to it? Do I need to include Qc in the work done (WD=Qh-Qc)?
Thanks for any help :)

2. Aug 8, 2012

### Simon Bridge

That's what I'd have done ... though I'd have explicitly stated the assumption of a Carnot cycle. I suppose I can see four marks in there: 1 each for the three ideas represented by the equations, and 1 for getting the math right. But then I tend to give whole marks for things.

note ...
mgh = (1000kg)(9.81N/kg)(3m) = 29430J ... 10x what you got?

3. Aug 8, 2012

### AGNuke

Sometimes 4 marks questions are by far the simplest questions of the paper. I am in 12th grade, and I am like

"Derive Mayer's Equation" - 3 marks (which takes a whole page if you aim to get full marks)

and solving an itty bitty question USING Mayer's Formula, not even two lines' length - 5 marks.
And they say we have superpower solving half the paper in last 10 minutes.

4. Aug 8, 2012

### Simon Bridge

As someone who writes exams...
A lot depends on what the person who wrote the marking schedule thinks about what you will find easy and what you should be rewarded for knowing/doing.

eg. deriving Mayer's relation should take about a dozen or so lines and comes in three parts, and is probably not as important as being able to apply it. Especially if the derivation was demonstrated in class and you were expected to memorize it. It'd be worth more if it were given, cold, in an assignment.

So that sort of thing makes sense to me ;)

5. Aug 9, 2012

### twinklestar28

Hi Simon

Sorry the mass is actually 100kg! not 1000kg so the energy would be 2943J but Im not sure if I should divide by 20 to get 147.15J per cycle before I plug it into the equation, should I plug in 2943J or 147.15J as the work done or should i divide by 20 after?

6. Aug 9, 2012

### Simon Bridge

Does it matter?

7. Aug 14, 2012

### twinklestar28

No it doesnt i was overthinking it i didnt check to see if its the same thing, sorry! thanks for the help :)

8. Aug 14, 2012

### Simon Bridge

In general - the same work may not be done on every cycle.
In which case, calculating the total work and dividing gives you the mean work per cycle.
Which is why I did it that way :) covers my ***.

<sigh> bits wot benefit from cover... donkey!