Three objects of same heat capacity with temperature T1=200 K, T2=400 K and T3=400 K exchange heat with each other. They are isolated from the rest of the universe. Find the highest possible temperature one of them can reach in kelvin.

Answer 1

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.