# S2 - S1 > Integral 1-2 dQ/T Irreversible Process. What T ?

1. May 12, 2013

### morrobay

For an irreversible process: S2 - S1 > ∫12 dQ/T
In the case of heat transfer dQ between object with T1 to object with T2, T1 > T2
ΔS in Joules/K is known But for evaluating ∫12 dQ/T = ln dQ/T - ln dQ/T . d/Q is known but what values for T in this case ?
Object 1 & 2 have Temperatures before heat transfer and different temperature after

2. May 13, 2013

### StarsRuler

The temperature is variable along the integration, while heat change. Only with the 2 temperatures extremis is not possible use these formula to solve any problem. You need an equation for heat transfer, for example. This is really no thermostatics, it requires thermodynamics, from empirical equations or from statistical mechanics treatments

3. May 13, 2013

### DrDu

morrobay, you are right that this formula has quite a restricted range of applicability as T may not be defined during a general irreversible process.

4. May 13, 2013

### morrobay

Thanks, would you please check my numerical problem on this in the Homework section;
Introductory Physics. Topic : Entropy for Irreversible Process.

5. May 13, 2013

### Andrew Mason

This is incorrect if by dQ you mean the actual heat flow in the process. To calculate the change in entropy, you have to use a reversible path between the beginning and end states

$$\Delta S = \int_{rev path} \frac{dQ_{rev}}{T}$$

AM

6. May 14, 2013

### morrobay

Yes it seems that S2-S1 > ∫12dQ/T
is just qualitative. As Chestermiller showed one way to quantify S2-S1
in heat transfer between hotter and cooler object is with.:
Δ S = mCv(ln T2/T1+lnT2/T1)
Where the first ln T2/T1 is for hotter object losing heat and second
ln T2/T1 for cooler object gaining heat.
What is the original integral that this evaluation was derived from ?

Last edited: May 14, 2013
7. May 14, 2013

### Andrew Mason

It is derived from ΔS = ∫dQrev/T:

For the hotter object (body A), a reversible path from T1 to T2 would be found by placing it in thermal contact with a body that has an infinitessimally lower temperature and whose temperature keeps decreasing so as to maintain that infinitessimal lower temperature as heat flows out of A, until it reaches T2. For that path:

$$\Delta S_A = \int_{T_1}^{T_2}\frac{dQ}{T}= \int_{T_1}^{T_2}\frac{mCdT}{T} = mC\ln T_2 - mC\ln T_1 = mC\ln\left(\frac{T_2}{T_1}\right)$$

Similarly the change in entropy of the cooler body, B is:

$$\Delta S_B = mC\ln\left(\frac{T_f}{T_i}\right)$$ where i and f denote initial and final temperatures of Body B.

AM

Last edited: May 14, 2013