Answer
a) 287 J
b) 38.5%
c) 363.94 K
d) 5,338.2 W
Work Step by Step
a) We find that the amount of work done is:
$ W = 745-458=287 \ J$
b) We know that the efficiency is equal to:
$=100(1-\frac{458}{745})=38.5 $%
c) We multiply the temperature of the warmer reservoir by the efficiency to find:
$T=(592)(1-.385)=363.94 K$
d) The power output is equal to the work done each cycle times the number of cycles per second, for Power is Joules per second. Thus, it follows:
$P = 287 \times 18.6 = 5,338.2 \ W$
