1 kg of iron with the temperature of 323 K to 293 K is cooled by dipping into a large water bath temperature of 293 K. What is the total entropy change? 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?!