the total amount of the work that we can get out is just given by the area inside this curve.

能够输出的功就是,我们可以得出来,曲线所包围的面积。

麻省理工公开课 - 热力学与动力学课程节选

So the maximum work out required the maximum heat in.

因此输出最大的功要求,有最大的热。

麻省理工公开课 - 热力学与动力学课程节选

So use work out of right-hand side to run left-hand backward.

利用右边输出的功,来驱动左边的热机反方向运行。

麻省理工公开课 - 热力学与动力学课程节选

The more work we can get out of it.

可以输出的功就越多。

麻省理工公开课 - 热力学与动力学课程节选

So delta u is just equal to the work but we also know what happens T2 because the temperature is changing from T1 to T2.

所以Δu等于输出的功,但我们也知道它会发生,同时我们知道温度从T1变到。

麻省理工公开课 - 热力学与动力学课程节选

It makes sense, right, because you know we got less work out and delta u is the same right, so it must be that less heat got transferred.

这是显而易见的，因为输出的功更少,且Δu相等，所以需要的热量更少。

麻省理工公开课 - 热力学与动力学课程节选