## Essential University Physics: Volume 1 (3rd Edition)

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: $P = 287 \times 18.6 = 5,338.2 \ W$