- #1

- 101

- 8

**Relevant equations**

dS = mcdT/T

S=mc *ln(T2/T1)

C iron = 450J/(kgK)

C water= 4200 J/(kgK)

m water = I see in my book 1 kg in solution

**The attempt at a solution**

I see in my book that they solved this problem like this :

**ΔS for Iron = mc ln(T2/T1)= 1*450* ln(293/323) (J/K)= -43.9 J/K**

**I dont have problem to understand this but I see for water that they did like this :**

**ΔS for water = mc*(T2-T1)/T1 = 1*450*30/(293) = 46.1 J/K**

**Here they used C iron to calculate**

**ΔS for water!**And the total increase of entropy in this system is

**46.1-43.9 = 2.2J/K**

I have two questions

*1)why can not we use integrals to find the entropy change in the water as we did for the iron ?*

2)and the second question is that why we use the specific heat capacity of iron when we'd calculate the entropy change in the water?!

2)and the second question is that why we use the specific heat capacity of iron when we'd calculate the entropy change in the water?!