0.10 L of water is slowly heated from 27 ℃ to 87 ℃. What is the...

Question

Question

0.10 L of water is slowly heated from 27 ℃ to 87 ℃. What is the change in entropy?

Answer 1

Answer #1

we knew given that

T1=273+27=300 K

T2=273+87=360K

m =density x volume

m= 0.1 x 1 =0.1 Kg

Change in entropy,

dS =m Cp In(T2/T1)

dS =0.1x4186 x In(360/300 dS =76.31 J/K

Answer 2

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