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His I. Q is 75. Well, we're all different, Mr. Hancock.
VOA: standard.other
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qB This must be heat q B, right. So again, but now let's also look at our special function.
这个不是，这应该是热量，对吧？所以又一次，这些是我们熟悉的。
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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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Let me give you, I'm going to create, q a polar point, I'm going to call it q, and we'll give it some random values.
让我给大家讲解下，我要去创建，一个极坐标点，然后我会去命名它为，然后我给它赋一些随机的值，好，现在我想知道。
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I am going to highlight this by calling this q1, lower case q being the charge here on the nucleus.
我要把它称作q1来进行强调，小写q是指原子核上的电量。
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And let me use a big Q to be the total quantity produced-- sorry--the total quantity demanded in the market.
用Q代表总产量,对不起,是市场总需求量
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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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U OK, so let's look at delta u. Delta u is q plus w.
好的，让我们看看Δ
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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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w And that has to equal q plus w, summed up for all the steps.
等于q加，对所有过程相加。
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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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w Now, we have u, q and w.
现在我们有u，q和。
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So just to emphasize that the strategies are quantities, rather than using S let me use Q today to be the strategies.
为了突出策略就是商品的产量,这次我们用Q来表示策略而不用S了
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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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Sorry, the question was, what happens if I said p is less than q?
抱歉，问题是，如果我这里写p小于q的话？
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Q So it's just capital W over capital Q, q1 which is to say it's minus all that stuff over q1.
所以效率就是W除以，等于负的所有这些之和除以。
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All right, q has to be just the opposite of this because we've already figured out that there's no change in u.
好，q一定与它,相差一个符号，因为我们已经,指出U是不变的。
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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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du We can talk about du for the system well, w that's q plus w.
我们可以讨论它的，等于q加。
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Adiabatic q equal to zero. It's also delta H 0 which is zero. The two didn't necessarily follow because remember, delta H is dq so p is only true for a reversible constant pressure process.
在这个过程中ΔH等于,绝热的所以q等于0，而ΔH也等于,这两个也不一定有因果关系,因为，记住，ΔH等于dq只有在恒压。
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Cp And delta T is given by the heat, which has to do with how much of the candle burnt, divided by the constant pressure heat capacity.
T等于热量q除以恒定的等压热容，其中热量与，蜡烛燃烧的多少相关。
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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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All right, so that's just solving out the algebra, so we're saying what q solves, a - c over 2b - q2 over 2 = 0.
没错,只需要解出当,/2b-q2/2=0,时q2的值即可
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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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w That's u2 minus u1, and it's q plus w.
就是u2减去u1，等于q加。
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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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All right, what we've already got that q is zero.
好，现在我们已经知道了q是零。
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So qi and q-i or q1 and q2 will be the strategies.
用qi,q-i,q1,q2表示策略
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