# Entropy changes while mixing liquids of different temperatures

If I mix cold milk with hot coffee in an insulated styrofoam cup, will the entropy of the coffee decreases by the same amount that the entropy of the milk increased?

• Once you mix the milk and coffee, the milk and coffee are no longer separate entities. And you also have to take into account the entropy of mixing (concentration changes). Do you really want to include that? Jan 1, 2019 at 14:28

No, it won't. Entropy will be created. Yes, the coffee will loose some entropy, but the milk will receive more entropy than the coffee will loose.

To be precise, entropy transfer follows an inverse temperature law

$$dS = \frac{dE}{T}$$

where $$T$$ is the temperature of the system that's receiving/loosing the energy/entropy. This temperature is different for the coffee and the milk, so $$dS_{milk}$$ will be greater than $$-dS_{coffee}$$:

$$dE_{milk} = -dE_{coffee}$$ $$T_{milk} < T_{coffee}$$ $$\Leftrightarrow \frac{1}{t_{milk}} > \frac{1}{T_{coffee}}$$ $$\Leftrightarrow \frac{dE_{milk}}{t_{milk}} > -\frac{dE_{coffee}}{T_{coffee}}$$ $$\Leftrightarrow dS_{milk} > -dS_{coffee}$$

• An explanation for the downvote would be nice... Jan 1, 2019 at 16:42

If you specify that the milk and coffee are brought into direct contact, but not allowed to mix intimately, then the first step in determining the entropy change of the system is to determine the final temperature $$T_f$$ of the system using the first law of thermodynamics:
$$m_cC_c(T_{c, init}-T_{f})=m_mC_m(T_f-T_{m, init})$$ This leads to $$T_f=\frac{m_cC_cT_{c, init}+m_mC_mT_{m, init}}{m_cC_c+m_mC_m}\tag{1}$$ The next step is devise separate (alternative) reversible processes for calculating the change in entropy of the coffee between its initial and final states and, separately, the change in entropy of the milk between its initial and final states. This leads to: $$\Delta S_c=m_cC_c\ln{\left(\frac{T_f}{T_{c,init}}\right)}$$ $$\Delta S_m=m_mC_m\ln{\left(\frac{T_f}{T_{m,init}}\right)}$$ The overall change in entropy for the combined system of coffee and milk is the sum of these: $$\Delta S=m_cC_c\ln{\left(\frac{T_f}{T_{c,init}}\right)}+m_mC_m\ln{\left(\frac{T_f}{T_{m,init}}\right)}\tag{2}$$ If you combine Eqns. 1 and 2 properly, you can show that $$\Delta S>0$$ for all combinations of non-equal values for the initial temperatures.