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His I. Q is 75. Well, we're all different, Mr. Hancock.
VOA: standard.other
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u=q+w All right, what is delta u? delta u is q plus w.
好，Δu是多少？Δ
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And of course, in either case, w delta u is q plus w, so it's q irreversible plus w irreversible, u being a state function it's the same in either case.
当然了，在这两个过程中，Δu都等于q加，因此现在是q不可逆加w不可逆，同时u在这两个过程中都是态函数。
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Delta u is q plus w. Delta u isn't zero.
U是q+w，△U不是零。
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For isothermal expansion, that means that delta u does not change, but delta q is equal to delta w?
在等温过程中，是不是内能不变，Δq等于Δw？
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q1 The heat input is just q1, Q and we'll define that as capital Q.
输入的热是，我们把它定义为大写的。
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Going from two to three, that's an adiabatic expansion, so q is equal to zero in that step.
从第二点到第三点，是绝热膨胀，因此q等于零。
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w This is just q plus w. There's w, RT1 ln q has to be R T1 log of V2 over V1.
而U等于q加，那是w，q应该是。
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It's well insulated. Heat is not going in or out adiabatic. q is equal to zero.
绝热性很好，热量不会变化,是绝热的，Q等于零。
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H=qp The u plus p V. Delta H is equal to q V.
括号里面的就是H，等于u+pv，Δ
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q Delta u is w2 prime.
等于零，q，is，zero。,Δu等于w2一撇。
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Or I could have a non-adiabatic, I could take the same temperature change, by taking a flame, or a heat source and heating up my substance. So, clearly q is going to depend on the path.
也能改变温度，绝热指的是没有热传递，在非绝热条件下，也同样可以升温,比如用火或者热源加热，这样，q也应当与路径有关。
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