It's so hot

For heating (or cooling) of any system (gas, liquid or solid) at constant pressure from an initial temperature $T_0$ to a final temperature , the entropy change is

$ΔS = nC_p \log \frac{T}{T_0}$ , where $n =$ number of object , $C_p =$ heat capacity, $T_0 =$ initial temperature, $T =$ Final temperature, then the total entropy of the objects is:

$ΔS= 3S_0 + C [\log(\frac{T_1}{T_0}) + \log(\frac{T_2}{T_0}) + \log(\frac{T_3}{T_0})$

$ΔS= 3S_0 + C[\log(\frac{200}{T_0}) + \log(\frac{400}{T_0}) + \log(\frac{400}{T_0})]$

but $ΔS$ not needed for this case , all we need to do to maximize $T$

That is easy to see, but typically the energy of the system will change. e.g. You can attempt to keep Energy constant but you will always do work on the system, so we keeping S and E constant implies that:

$T_1+T_2+T_3 = 1000K$

$a+b+b = 1000K$

$2a = b$ $a = \frac{1}{2} b$

$T_1. T_2. T_3 = 200 \times 400 \times 400 = 32000000K$

$ab^2 = 32000000$ also $a + 2b = 200 + 400 + 400 = 1000$ ,

Then, from both equation we obtain $a(500 - \frac{a}{2})^2 = 32000000$ , solve the equation we obtain that :

$a = 200$ , $487.69$ , and $1312.3$ . Since the problem ask us, to find maximum value, then if $a = 1312. 3$ , it will not satisfy, because it will implies $b < 0$ , then $a = 487.69$ , and satisfy both equation. So,

the highest value that we can reach is $\boxed{487.69}$

Suppose when the objects are at a temperature $T_0$ , each has an entropy of $S_0$ . The total entropy of the system is given by

$S_{total}=3S_0+C(ln \frac{T_1}{T_0} + ln \frac{T_2}{T_0} + ln \frac{T_3}{T_0} )$ , where C denotes the heat capacity of the objects.

In an isolated system, the internal energy is conserved and the total entropy does not decrease.

We have, $\Delta U= E_0 + C(T_1 + T_2 + T_3)$

Hence, $T_1+T_2+T_3=1000K$

In order to find the maximum temperature while keeping the entropy constant at the same time, $T_1*T_2*T_3=200\times400\times400=3.2\times 10^7K^3$

Suppose one of the objects can reach the highest temperature $T_3$ , the other two objects would have the same temperature $T_1=T_2$ , therefore $2T_1+T_3=1000$

Now we obtain an equation,

$T_3(500-\frac{1}{2} T_3)^2=3.2\times10^7$

Solving this cubic equation which yields three solutions, $T_3 = 200, 487.7, 1312$

The answer cannot be $200K$ because it gives a higher temperature for the other two objects, $1312K$ is rejected as well since the other two objects cannot have negative temperature.

Hence the answer is $487.7K$ .

From energy conservation (since they have the same heat capacity): $t_1 + t_2 + t_3 = T_1 + T_2 + T_3$ (1)

Entropy change ( $C_1=C_2=C_3=C$ ): $-C \ln(\frac{t_1}{T_1}) - C\ln(\frac{t_2}{T_2}) - C \ln(\frac{t_3}{T_3}) = \ln(\frac{T_1^{C} T_2^{C} T_3^{C}}{t_1^{C} t_2^{C} t_3^{C}})=\Delta S \geq 0 \Rightarrow t_1 t_2 t_3 \leq T_1 T_2 T_3$

Choose the upper condition $t_1 t_2 t_3 = T_1 T_2 T_3$ (2).

Now let's find the highest possible temperature for the first object. From (2) one can write $t_2 = (\frac{T_1 T_2 T_3}{t_1 t_3})$ and plug it in (1):

$t_1 + t_2 + t_3 = T_1 + T_2 + T_3 \Rightarrow \frac{d}{dt_3} t_1 + \frac{d}{dt_3} t_2 + 1= 0$

$\Rightarrow (t_1)' - (T_1 T_2 T_3) (t_3^{-1} t_1^{-2}(t_1)'+ t_1^{-1} t_3^{-2}) + 1 = 0$

At extreme value, $(t_1)' = \frac{d}{dt_3} t_1 = 0$ , so one arrives at $t_3 = \sqrt{\frac{T_1 T_2 T_3}{t_1}}$ . Also, $t_2 = (\frac{T_1 T_2 T_3}{t_1 t_3}) = \sqrt{\frac{T_1 T_2 T_3}{t_1}} = t_3$

Finally, put it in (1) to get (for maximum $t_1$ ):

$t_1 + t_2 + t_3 = t_1 + 2 \sqrt{\frac{T_1 T_2 T_3}{t_1}} = T_1 + T_2 + T_3 \Rightarrow t_1 = (9-\sqrt{17}) 100 = 487.69~\mbox{K}$

For others, we also arrive at the same equation by symmetry, so that the highest possible temperature is $487.69~\mbox{K}$ .

If the entropy of the objects at some T0 is S0 then the total entropy of the objects is:

S(T1,T2,T3) = 3 S0 + C [Log(T1/T0) + Log(T2/T0) + Log(T3/T0)]

We choose T0 such that the heat capacity can be taken to be constant for T > T0. Note that the heat capacity of any system must tend to zero for T to zero, so it would be incorrect to write the entropy as the c times the sum of the logarithms of the temperatures, even though it would give the right answer to this problem (and that's because the answer doesn't depend on T0 and S0).

When heat is exchanged bwtwen the systems, S canot decrease while the internal energy, given by:

E = E0 + C (T1 + T2 + T3 - 3 T0)

stays constant.

To find the highest possible temperature you need to maximize one of the temperatures, say T1, while keeping E and S constant. Keeping E constant means that:

T1 + T2 + T3 = const. = 1000 K

Keeping S constant implies that:

T1 T2 T3 = const. = 32*10^6 K^3

Or:

Log(T1) + Log(T2) + Log(T3) =

Log(32*10^6 K^3)

Maximize T1 using Lagrange multipliers:

1 = lambda1 + lambda2/T1 (1)

0 = lambda1 + lambda2/T2 (2)

0 = lambda1 + lambda2/T3 (3)

From (2) and (3) we see that T2 = T3.

We then have that:

T1 + T2 + T3 = T1 + 2 T2 = 1000 K

T1 T2^2 = 32*10^6 K^3

So:

T1 (500K - T1/2)^2 = 32*10^6 K^3

T1 (1000 K - T1)^2 - 128*10^6 K^3 = 0

T1 = 200 K has to be a solution (but obviously not the optimal one) given the initial temperatures, so we put T1 = (200 + x)K to get a quadratic equation:

(200 + x) (800 - x)^2 - 128*10^6 = 0

200 x - 32000 + 640000 - 1600 x + x^2

x^2 -1400 x + 320000 = 0 ------>

x = 700 +/- 100 sqrt(17)

This gives T1 = 487.689 K and T1 = 1312.31 as solutions. The latter yields negative temperatures for the two other objects, so the highest possible temperature is 487.689 K.