4) How many g of water at 25.0 degree C must be added to a Dewar flask containin

Question

4) How many g of water at 25.0 degree C must be added to a Dewar flask containing 20.0 g of ice at -5.0 degree C to satisfy the conditions that the final temperature inside the Dewar flask is 0 degree C and half the ice melts. Compute the entropy change to go from the initial to final state inside the Dewar. This is a constant pressure system. Use the following information in your calculations: the constant pressure heat capacity of liquid water = 4.18 J/g-K. the constant pressure heat capacity of ice = 2.09 Jig-K, and the enthalpy of fusion for water/ice = 334 J/g.

Explanation / Answer

4)
Heat required to take ice from -5 oC()268 K to 0 oC (273 K)
Q1 = m*Ci*delta T
      = 20*(2.09)*(5)
      = 209 J
Entropy change:
delta S1 = Cp ln (T2/T1)
                 = 2.09 * ln (273/268)  
                 = 0.039 J/K

Heat required to melt half og ice:
Q2= m*L
      = 10 * 334
      =3340 J

entropy change:
delta S2 = Q2/T
                 = 3340/273
                 =12.23 J/K

Let mass of water be m g
Heat required by water:
Q3= m*C*delta T
      = m*4.18*25
      = 104.5 m J

balance heat,
Q1+Q2 = Q3
209 + 3340 = 104.5*m
m=34 g
Mass of water required = 34 g

entropy change for water
delta S3 = Cp* ln (T2/T1)
                   = 4.18*ln (273/298)
                   = 0.37 J/K

total entropy change:
delta S = delta S1 + delta S2 + delta S3
               = 0.039 + 12.23 + 0.37
               = 12.64 J/K

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