Calculate the total change in entropy if 2.55 kg of wat 17.5 degrees Celcius is

Question

Calculate the total change in entropy if 2.55 kg of wat 17.5 degrees Celcius is mixed with another 5.75 kg of water at 67.5 degrees Celcius. The specific heat of water at contant pressure is 4186 J/(kg (K))

J/K

Okay if anyone can check my math and make sure I am not doing anything wrong I did the math and sapling , the homework site I used for physics tells me I am incorrect and I have no idea on any other ways to solve it, but this is what I did:

m1 x c1x (t-t1)= m2 x c2x(t2-t)

2.55 x (t-17.5)= 5.75x(67.5-t)

2.55t-44.625=388.125-5.75t

8.3.t=432.75

t= 52 degrees

T=273+52=325 K

T1= 273+17.5=290.5 K

T2= 273+67.5=340.5 K

S1= (2.55 K) (4186 J/(kg (k)) (ln (325/290.5)= 1197.88 J/K

S2= (5.75 kg) (4186 J/ (kg (k)) (ln(325/340.5))= -1121.39 J/K

S= S1+S2

S= 1197.88 J/K + (-1121.39 J/K)= 76.49 J/K <--- Sapling says this is incorrect I am very confused any help would be greatly appreciated.

Explanation / Answer

S = Cp * ln(Tfinal/Tinitial)

Tf = 52.14 = 325.29 ( 273 .15 K = 0c )

S1 = 2.55 x 4186 * ln( 325.29/290.65) = 1201.897 J/K

S2 = 5.75 x 4186 * ln( 325.29/340.65) = -1110.53 J/K

S = S1 + S2 = 91.367 J/K

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