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They may have a high body temperature and some areas of skin may turn red.
VOA: special.2009.05.26
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Yet, the temperature goes up. So, I can have a temperature change which is an adiabatic temperature change.
它与外界不会，有物质或者能量的流动,然而系统的温度升高了。
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And temperature then is associated with property And if it had changed, then the temperature between those two would have changed in a very particular way.
就与这种性质有关,如果它发生了改变，说明这,两者的温度,以特定的方式发生了改变。
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So now we have a constant volume reversible temperature change.
所以现在我们有一个,等体，可逆的温度变化。
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I want to cool a gas with a Joule-Thomson experiment, what temperature do I have to be at?
给气体降温时,需要到达什么温度？
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The same temperature increase, work and heat, and we have a relationship between heat and work.
同样的温度升高，功和热，因此我们可以得到功和热的关系了。
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All right, so now we have the makings of a good thermometer and a good temperature scale.
这一常数只决定于温度,于是我们现在,可以定义一个理想的温度计和温标了。
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Now, you can have a change of temperature without any heat being involved.
不进行热量传递也可以改变温度,想象有一个热绝缘的盒子。
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You could take a continuous path, where you have an infinite equilibrium points in between the two, a smooth path, whereyoudrop pressureand temperature simultaneosly in little increments.
你可以选择光滑的连续路径，在初末态间有,无数个平衡态的点，压强和温度同时一点,一点地下降。
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This is real, unlike the Joule coefficient which is very small so that most gases have tiny Joule coefficients. So if you do a Joule experiment, you hardly measure a temperature change. With real gases, here you do actually measure it. You can feel it with your finger on your bicycle tire.
系数那样小以至于,大多数气体的焦耳系数,都很小，所以如果你做焦耳实验,很难测量出温度的变化,对于真实气体，你可以测量它，你能通过手指按在,自行车轮胎上来感觉到它。
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If I look at different points in my container during that path, I'm going to have to use a different value of pressure or different value of temperature That's not an equilibrium state, and that process turns out then to be an irreversible process.
如果我要研究在路径中容器里的,不同的点，我就得在容器里不同的点上使用,不同的压强值,或不同的温度值,实际上这不是个平衡态，这个过程是,不可逆过程。
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It's just how much heat is involved when we change the temperature. Now, the products have some heat capacity associated with them right, it takes a certain amount of heat if we make their temperature change, to either put it in or take it away, depending on which direction the temperature is changing.
问题就是当我们改变温度时,有多少热量发生了转移,生成物具有一定的热容,如果我们改变,它们的温度，就要输入或,提取一定的热量,这取决于温度改变的方向。
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We have an interpolation scheme between zero and 273.16 with two values for this quantity, and we have a linear interpolation that defines our temperature scale, our Kelvin temperature scale.
的两个值做线性插值,就得到了开尔文温标,直线的斜率等于水的三相点,也就是这一点处的f的值,再除以273。16，这是这条直线的斜率，这个量，f在三相点处的值。
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V They do have the same temperature though.
不同，a，different，volume。,但是温度是一样的。
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And so, we have this cartoon. Again, we have an open beaker and a candle, and we're putting only heat into this beaker, T2 and the temperature goes from T1 to T2.
好的，我们要再次利用这幅图，这有一个敞开的烧杯和一根蜡烛，我们对烧杯加热，温度从T1上升到。
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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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