You want to design an ideal Carnot heat engine that wastes only 35.0% of the heat that goes into it. The lowest cold-reservoir temperature available to you is +15.0°C. If 150.0 J of work is done per cycle, the heat input per cycle is closest to

The heat input per cycle is closest to 230.8 J

Explanation:

Given;

percentage of energy waste = 35%

The lowest cold-reservoir temperature = 15.0°C

useful work done in the reservoir per cycle = 150.0 J

To design an ideal Carnot heat engine that wastes only 35.0% of the heat that goes into it.

If 35% of heat input = wasted work, then

65 % of heat input  = useful work

∴ 0.65 ×  heat input = 150

heat input = 150/0.65

heat input = 230.8 J

Therefore, the heat input per cycle is closest to 230.8 J

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.