# Homework Help: Exergy value at double temperature

1. Sep 21, 2014

### Maylis

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?

2. Sep 21, 2014

### rude man

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?

3. Sep 21, 2014

### Maylis

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?

4. Sep 22, 2014

### rude man

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.