[confusedstud]

When I am trying to prove that entropy change is positive during an irreversible heating process, we have to use this proof http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node41.html in the hot brick process.

But what if my two bricks had different C values? How would I proof it then?

Also, in this hot brick case it would be an irreversible process right? But in the case where two reservoirs of heat are placed beside each other would it be considered a reversible process or irreversible process?

My notes states that the entropy of the universe will increase for an irreversible process while it remains constant for a reversible process. But in the case of two heat reservoirs the total entropy will increase so it seems like a irreversible process. So are both cases irreversible processes but we are using the reversible path to calculate the entropy change of an irreversible path?

[mjc123]

But what if my two bricks had different C values? How would I proof it then?

Just go through the maths with two different C values, C₁ and C₂.
From the form of the expressions it is clear that T_M is the arithmetic mean of T_H and T_C (weighted by the C values), while for ΔS to be zero, T_M would have to be the weighted geometric mean (e.g. in the given example, if T_M = sqrt(T_HT_C), the quantity in brackets is 1, so its log is zero). But the arithmetic mean is always greater than the geometric mean, so ΔS > 0.

[confusedstud]

But what if my two bricks had different C values? How would I proof it then?

Just go through the maths with two different C values, C₁ and C₂.
From the form of the expressions it is clear that T_M is the arithmetic mean of T_H and T_C (weighted by the C values), while for ΔS to be zero, T_M would have to be the weighted geometric mean (e.g. in the given example, if T_M = sqrt(T_HT_C), the quantity in brackets is 1, so its log is zero). But the arithmetic mean is always greater than the geometric mean, so ΔS > 0.

Could you show that? Sorry my mathematics is not very strong so I can't really understand it.

On a similar note for my second question whereby two heat reservoirs are placed side by side. Would this be classified as a reversible or irreversible process? I think its an irreversible process as the total entropy increases but we can use the reversible path's calculations for it. Is this correct?

[mjc123]

Q_cold = -Q_hot
C_c(T_m - T_c) = C_h(T_h - T_m)
T_m = (C_hT_h + C_cT_c)/(C_h + C_c)
ΔS_cold = C_cln(T_m/T_c) = C_c(lnT_m - lnT_c)
ΔS_hot = C_hln(T_m/T_h) = C_h(lnT_m - lnT_h)
Hence ΔS_total = 0 if lnT_m = (C_hlnT_h + C_clnT_c)/(C_h + C_c)
i.e. if lnT_m is the weighted average of lnT_c and lnT_h, which means T_m is the weighted geometric mean of T_c and T_h.
But it is really the weighted arithmetic mean, as we saw above. And the arithmetic mean is always greater than the geometric mean. So T_m is higher than the value that would give zero entropy change, which means the hot brick loses less entropy and the cold brick gains more.
Think about it graphically. If you have a point half way between lnA and lnB on a log scale, where is the equivalent point going to come between A and B on a linear scale? Half way? Or more, or less?