Exergy value at double temperature

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gfd43tg
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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.