Compare final T of reversible/irreversible equations

[accountkiller]

Homework Statement

Suppose C_A = C_B = C in equations 2.93 and 2.109 (shown below). Compare the form of expressions for the final temperature.

Homework Equations

(2.93) T = \frac{C_{A}T_{A} + C_{B}T_{B}}{C_{A} + C_{B}}
(2.109) T = T_{A}^{\frac{C_{A}}{(C_{A}+C_{B})}} T_{B}^{\frac{C_{B}}{(C_{A}+C_{B})}}

(2.93) is for an irreversible process
(2.109) is for a reversible process
T is the final equilibrium temperature when two bodies are put into thermal contact.
C is specific heat capacity

The Attempt at a Solution

The problem is that I don't know where to start. I'm told in the sentence before equation (2.93) that in that equation, T_A > T_B. Other than that, I'm not sure exactly how or what I'm supposed to be deducing about the final temperature just from looking at the two equations. I'm told the answer is that the final temperature will be T^A but have no idea why. What's my first step? How should I start thinking about it?

I'd appreciate any guidance. Thanks :)

 

[mstud]

I'd say that...
If we suppose C_A = C_B = C, as you're told to, start by just replacing both C_A and C_B with C, and simplify the formulas:

(2.93) T = \frac{C T_{A} + CT_{B}}{2C}= \frac{C( T_{A} + T_{B})}{2C}= \frac{ T_{A} + T_{B}}{C}.

(2.109) T = T_{A}^{\frac{C}{(2C)}} T_{B}^{\frac{C}{(2C)}}=T_{A} \ ^{\frac{1}{C}} \cdot T_{B} \ ^{\frac{1}{C}}

Do you have any further ideas now ?

Homework Statement

Suppose C_A = C_B = C in equations 2.93 and 2.109 (shown below). Compare the form of expressions for the final temperature.

Homework Equations

(2.93) T = \frac{C_{A}T_{A} + C_{B}T_{B}}{C_{A} + C_{B}}
(2.109) T = T_{A}^{\frac{C_{A}}{(C_{A}+C_{B})}} T_{B}^{\frac{C_{B}}{(C_{A}+C_{B})}}

(2.93) is for an irreversible process
(2.109) is for a reversible process
T is the final equilibrium temperature when two bodies are put into thermal contact.
C is specific heat capacity

The Attempt at a Solution

The problem is that I don't know where to start. I'm told in the sentence before equation (2.93) that in that equation, T_A > T_B. Other than that, I'm not sure exactly how or what I'm supposed to be deducing about the final temperature just from looking at the two equations. I'm told the answer is that the final temperature will be T^A but have no idea why. What's my first step? How should I start thinking about it?

I'd appreciate any guidance. Thanks :)

 

[accountkiller]

I'd say that...
If we suppose C_A = C_B = C, as you're told to, start by just replacing both C_A and C_B with C, and simplify the formulas:

(2.93) T = \frac{C T_{A} + CT_{B}}{2C}= \frac{C( T_{A} + T_{B})}{2C}= \frac{ T_{A} + T_{B}}{C}.

(2.109) T = T_{A}^{\frac{C}{(2C)}} T_{B}^{\frac{C}{(2C)}}=T_{A} \ ^{\frac{1}{C}} \cdot T_{B} \ ^{\frac{1}{C}}

Do you have any further ideas now ?

If by \frac{T_{A} + T_{B}}{C}, you mean \frac{T_A} + T_{B}}{2} for the first one... and replace the C by 2 again in the second one, then yes I realize I can do that. I didn't think to use it here but I did use that in the next question in the homework assignment.. I suppose it would have made more sense to do it here, thanks.

But no, I don't have any further ideas from that. In the next question on the homework, we're given values for T_A and T_B and I find that plugging them into 2.93 yields a higher resulting value than plugging the same numbers into 2.109, so that tells me something. But without numbers and just looking at the equations above reduced so the C is gone... am I just supposed to be able to deduce from that that one process is higher than the other simply based on math?