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.
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.
To find the highest possible temperature one of the objects can reach, we need to consider the principle of energy conservation and the fact that heat always flows from a hotter object to a colder one until thermal equilibrium is reached.
In this scenario, Object 1 has the lowest temperature of T1 = 200 K, while Objects 2 and 3 have the same temperature of T2 = T3 = 400 K.
Since Object 1 is the coldest, it will absorb heat from the other two objects until it reaches thermal equilibrium. The amount of heat gained by Object 1 will be equal to the amount of heat lost by Object 2 and Object 3 combined.
Let's assume the heat capacity of each object is denoted by C. Therefore, the heat exchange equation can be written as:
C * ΔT1 = (2C) * ΔT
Here, ΔT1 is the change in temperature of Object 1 and ΔT is the change in temperature of Objects 2 and 3. Note that the heat capacity doubles for Objects 2 and 3 since there are two of them.
We can rearrange the equation to solve for ΔT1:
ΔT1 = 2 * ΔT
Now, using the equation for heat exchange:
C * ΔT1 = (2C) * ΔT
C * (2 * ΔT) = (2C) * ΔT
The heat capacities, C, cancel out from both sides of the equation, leaving us with:
2 * ΔT = ΔT
Since the change in temperature, ΔT, is positive, we can solve for it:
ΔT = 0
This means that the change in temperature for Objects 2 and 3, ΔT, is zero. Therefore, their temperatures will remain the same at T2 = T3 = 400 K.
Substituting this value back into the equation for ΔT1:
ΔT1 = 2 * ΔT
ΔT1 = 2 * 0
ΔT1 = 0
Since ΔT1 is also zero, it means that Object 1 will not experience any change in temperature. Therefore, the highest possible temperature that any of the objects can reach is still the initial temperature of 400 K for Objects 2 and 3.