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Delta u, delta H, familiar state functions, q w changes in their values, q, w, heat and work.
U，△H，很熟悉的态函数，它们的值在变化；
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In the meantime, I believe that the U.S. is doing quite a bit of work in terms of raising fuel efficiency standards,
同时，我相信美国在提高燃料效率标准方面做了不少事，
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q Work is zero. Delta u is w plus q, work plus heat. This is zero, this isn't.
功是零，△U是w加，功加上热量，这个是零。
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In the case of France again, they are the work of a brilliant military engineer called Vauban, v-a-u-b-a-n.
例如法国,在佩皮尼昂 里尔 蒙梅迪等地方,这样的要塞城镇随处可见
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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。
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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。
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U And then if we can also determine delta u, then we know this, we know delta u is q plus w w, then we can determine work as well, right?
然后，如果我们也能定出△,然后我们知道这个,我们知道△U等于q加？
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U It's u, because u is to q plus w right, heat and work, but it's adiabatic. So there's no heat, exchange with the environment, and it's constant volume, so there's no p dV work, right.
什么是零？是U，因为,等于q加w，热量和功，但这是绝热的，所以系统与环境间没有热量交换；,同时它是灯体的，所以也没有pdV形式的功。
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