Physics: Thermodynamics (ice)

Question

If a 0.06247 kg ice cube initially at 0.00 C is allowed to melt in a Styrofoam cup holding 0.497 kg of water at 10.0 C , what is the total entropy change for this process?

Answer 1

Hi,

 

I am happy to explain this to you:

 

First you should find the amount of energy transferred. This is done by using the heat of fusion:

 

Q= (mass)(latent heat of fusion)

Q= m*L_f

 

L_{f ice}= 333kJ/kg

L_{f H2O}= 334kJ/kg

 

so: Q_ice =.06247kg (333kJ/kg) = 20.8kJ and Q_H20= .497kg(334kJ/kg) 

 

Now entropy: S= (energy transferrred)/(temperature)

 

S=Q/T

 

Temperature is measured in kelvin

 

S_ice = 20.8kJ/273.15K = .07kJ/k

S_H20 = .586kJ/K

 

Now for the change in entropy we will take the entropy of water minus the entropy of ice

ΔS= .586kJ/k -.07kJ/K = .516 kJ/K

 

Thus the entropy increased and is consistent with the physical law that entropy either remains the same or increases!

I hope this helps!