The second question is that the first floor is wrong. Now the entropy change is calculated as follows:

1, and find the equilibrium temperature t after mixing.

Taking two liquids as a system, assuming that cp is constant, the expansion coefficient of the liquid has nothing to do with temperature, so there are

Q=0 and W=0.

So Mcp(T-T 1)+Mcp(T-T2)=0.

T=(T 1+T2)/2 is obtained.

2. Entropy change of liquid

According to the definition and meaning of the question, there is ds = Δ q/t = CP × dt/t.

So for the liquid with T 1, its entropy becomes ds1= m× CP× dt/t.

Its temperature ranges from T 1→(T 1+T2)/2, so the above integral is: △ s1= mcpln [(t1+t2)].

Similarly, △ S2 = mcpln [(t1+t2)/(2t2)]

Therefore, after mixing, the total entropy of the system becomes

△s=△s 1+△s2=m×cp×ln[(t 1+t2)^2/(4t 1×t2)]

Over.