Exergy value at double temperature

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Heat at 400°C does not carry double the exergy value of heat at 200°C, as demonstrated through calculations involving the lost work formula and exergy equations. The difference in exergy between the two temperatures is expressed as Exergy_{400} = -200 ΔS + Exergy_{200}, indicating a linear relationship rather than a doubling effect. Discussions also highlight the role of specific heat capacity (C) and its dependency on temperature, suggesting that the exact value of exergy is influenced by the system's characteristics. There is some uncertainty regarding the appropriate specific heat to use, whether it be C_p or C_V. Overall, the consensus is that exergy increases with temperature but does not double when comparing these two specific temperatures.
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HS2: Does heat at 400°C carry two times exergy value than heat a 200 °C? If not, how much should it be?

Exergy is the maximum useful work that can be achieved. The lost work of a process is

##W_{lost} = T_{\sigma}\Delta S - Q##

Where is the surrounding absolute temperature.

If the Exergy is ##W_{ideal} - W_{lost}##, then

##Exergy_{400} = W_{ideal} - (673 K)\Delta S + Q##
##Exergy_{200} = W_{ideal} - (473 K)\Delta S + Q##

Subtracting these equations,

##Exergy_{400} - Exergy_{200} = -200 \Delta S##
##Exergy_{400} = -200 \Delta S + Exergy_{200}##

Which is clearly not the same as a doubling in value. Is this the right way to do it?
 
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This is just my shot at it, look for others to respond also:

I don't know about your Wlost formula. Never seen it. But if the surroundings are at zero K, is the work lost = -Q? I would think zero instead.

Since the low temperature T2 can be zero K, all the heat extracted at T = T1 can go into work. So W = Q1. But Q1 = C*T1. So assuming C constant between 200C and 400C, what W can be extracted at each temperature?
 
So then you would say for 400C, then ##Exergy_{400} = C_{p}(673K)##, and ##Exergy_{200} = C_{p}(473 K)##, which still would not be double?
 
Maylis said:
So then you would say for 400C, then ##Exergy_{400} = C_{p}(673K)##, and ##Exergy_{200} = C_{p}(473 K)##, which still would not be double?

Yes, except I'm not sure if it would be ##C_{p}## or ##C_{V}## or something else. I guess it would depend on the system. I would just call it C. In reality C would be a function of temperature anyway, for such a large range of temperatures all the way down to zero K.
 
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