Answer:
a) This heat engine is possible since real efficiency is lesser than theoretical efficiency, b)
, c)
.
Step-by-step explanation:
a) The maximum theoretical efficiency of a heat efficiency is given by the Carnot's cycle, whose formula is:
![\eta_(max) = \left(1 - (773.15\,K)/(1473.15\,K)\right)* 100\,\%](https://img.qammunity.org/2021/formulas/engineering/college/58cefn0ul9gbdekvzbie1xt0f45h33zedx.png)
![\eta_(max) = 47.5\,\%](https://img.qammunity.org/2021/formulas/engineering/college/db3u4ww6voxbexxv8wkkxvuz2jm7i5pfxo.png)
This heat engine is possible since real efficiency is lesser than theoretical efficiency.
b) The heat supplied to the engine by the heat source:
![Q_(H) = (W)/(\eta_(real))](https://img.qammunity.org/2021/formulas/engineering/college/44nnzve2cem65sr2r2hb1hkjs60qzxfh1a.png)
![Q_(H) = (500\,kJ)/(0.4)](https://img.qammunity.org/2021/formulas/engineering/college/2ocik82nhld9tstpqk7e81l2bfl988fr6u.png)
![Q_(H) = 1250\,kJ](https://img.qammunity.org/2021/formulas/engineering/college/vl8v48m734ees7i9ydpbs2qprbds96uqbt.png)
c) The heat rejected to heat sink is:
![Q_(L) = Q_(H) - W](https://img.qammunity.org/2021/formulas/engineering/college/a2sop80sce6rdc90w89f0a190iu9msf8x4.png)
![Q_(L) = 1250\,kJ - 500\,kJ](https://img.qammunity.org/2021/formulas/engineering/college/ii0380jhcjeul9ztyaze5emajgt8ebiqup.png)
![Q_(L) = 750\,kJ](https://img.qammunity.org/2021/formulas/engineering/college/18onbe3tn4p8y61jzkvez2z3mrm2nqsfr1.png)