Question

75 g of water at  69 ∘C are mixed with 75 g of water at 20 ∘C in a thermally-insulated container.

Calculate the change in entropy, ΔS.

I know the answer is  ΔS = 1.87 J/K, but I cannot remember what I did to get this last week. Thank you!

Answer 1

Let Tf be the final temperature. Mass of each water = 75 g heat energy associated with hot water is hot = m* C* AT = 75 g * 4.184 J/(°C.g) * (Tf - 69) °C heat energy associated with hot water is cold = m*C* AT = 75 g * 4.184 J/(°C.g) * (Tf-20) °C At thermal equilibrium, ahot = -9cold 75 g * 4.184 J/(°C.g) * (Tf - 69) °C = -75 g * 4.184 J/(°C.g) * (Tf - 20) °C Tf-69 = - (Tf - 20) 2 Tf = 69+20 = 89 Tf = 89/2 = 44.5 °C Final temperature of the system = 44.5 °C Since entropy is a state function, it can be calculated by means of two processes. 1. cooling of hot water 2. heating of cool water.

Tf = 44.5 °C = 44.5 + 273.15 = 317.65 K

Hot temperature = 69 °C = 69 + 273.15 = 342.15 K

Cold temperature = 20 °C = 20 + 273.15 = 293.15 K

Specific heat capacity = 4.184 J/(°C.g) = 4.184 J/(K.g)

AS = ".x XComax on it and Co k tomu na 317.6 317.65 S = 75 g x 4.184 J/(K.g) x In + 75 gx 4.184 J/(°C.g)x In – 293.15) 317.65 317.65 AS = 75 g x 4.184 J/(K.g) x In + ln – 342.15 293.15 AS = 75 gx4.184 J/(K.g)x(-0.0743+0.080266) AS = 75 gx4.184 J/(K.g)x(0.005967)=1.87233 J/K AS = 1.87J/K The change in entropy = 1.87 J/K