#### Answer

a) The efficiency is about 2/3 of the maximum possible efficiency.
b) $2154\ MW$
c) $93,673 \ houses$

#### Work Step by Step

a) We find the maximum efficiency, being sure to use the temperatures in Kelvin:
$= 1-\frac{303K}{553K}\times 100=45$%
b) We first find the power that the plant is producing:
$=\frac{880}{.29}=3034\ MW$
We subtract the amount that it actually gets to find that the amount wasted is:
$= 3034-880=2154\ MW$
c) We find:
$=\frac{2154\times10^6\ W }{23\times10^3}=93,673 \ houses$