part a)
heat transfers to enviroment = (1 - efficiency) * heat input
heat transfers to enviroment = (1 - 0.38) * 1.5 *10^14
heat transfers to enviroment = 9.3 *1013 J
part b)
ratio of heat transfer to work output = 9.3 *10^13/(0.38 * 1.5 *10^14)
ratio of heat transfer to work output = 1.63
the ratio of heat transfer to work output is 1.63
part c)
work done = 0.38 * heat input
work done = 0.38 * 1.5 *10^14
work done = 5.7 *1013 J
(37) Problem 5: An electrical power station reques 1.5x 1014 of heat transfer into the one...