| Date | November 2016 | Marks available | 2 | Reference code | 16N.3.SL.TZ0.10 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Explain | Question number | 10 | Adapted from | N/A |
Question
An ideal nuclear power plant can be modelled as a heat engine that operates between a hot temperature of 612°C and a cold temperature of 349°C.
Calculate the Carnot efficiency of the nuclear power plant.
Explain, with a reason, why a real nuclear power plant operating between the stated temperatures cannot reach the efficiency calculated in (a).
The nuclear power plant works at 71.0% of the Carnot efficiency. The power produced is 1.33 GW. Calculate how much waste thermal energy is released per hour.
Discuss the production of waste heat by the power plant with reference to the first law and the second law of thermodynamics.
Markscheme
correct conversion to K «622 K cold, 885 K hot»
\({\eta _{{\rm{Carnot}}}} = 1 - \frac{{{T_{{\rm{cold}}}}}}{{{T_{{\rm{hot}}}}}} = 1 - \frac{{622}}{{885}} = 0.297\) or 29.7%
Award [1 max] if temperatures are not converted to K, giving result 0.430.
the Carnot efficiency is the maximum possible
the Carnot cycle is theoretical/reversible/impossible/infinitely slow
energy losses to surroundings «friction, electrical losses, heat losses, sound energy»
OWTTE
0.71 × 0.297 = 0.211
Allow solution utilizing wasted power «78.9%».
1.33/0.211 × 0.789 = 4.97 «GW»
4.97 GW × 3600 = 1.79 × 1013 «J»
Award [2 max] if 71% used as the overall efficiency giving an answer of 1.96 × 1012 J.
Award [3] for bald correct answer.
Watch for ECF from (a).
Law 1: net thermal energy flow is QIN–QOUT
QOUT refers to “waste heat”
Law 1: QIN–QOUT = ∆Q=∆W as ∆U is zero
Law 2: does not forbid QOUT=0
Law 2: no power plant can cover 100% of QIN into work
Law 2: total entropy must increase so some Q must enter surroundings
OWTTE
