Question

A 0.850 kg sample of water is initially ice at temperature –24°C. What is the sample's entropy change if its temperature is increased to 54°C?

Answer 1

We know that the entropy change

∆s1=ms Ln(T2/T1)

When the change of the ice temperature T1=-24°C =-24+273.15 =249.15 K to T2=0°=273.15 K

Here m=0.85 kg , s=2093 J/kg°C

∆s1 =0.85×2093×Ln(273.15/249.15)

∆s1 =163.61 J/K

When the ice change to water at 0°C =273.15K

∆s2 =mL/T

∆s2=0.85×3.34×10^5/273.15

∆s2=1039.36 J/K

When the change of the water temperature from 0°C to 54°C

Here m=0.85J/kg°C , S=4186 J/kg°C , T2=54°C =327.15K , T1 =0°C =273.15 K

∆S3 =mSLn(T2/T1)

∆S3 =0.85×4186×Ln(327.15/273.15)

∆S3=641.87J/K

∆S =∆S1+∆S2+∆S3

∆S=163.61+1039.36+641.87

∆S =1844.84 J/K. Answer.