A small piece of ice (mass = 0.10 kg) at an initial temperature
of 0oC is placed within glass of water, where the
water's mass is 0.40 kg and it is at an initial temperature of
30oC.
(a) What is the final temperature of the system when thermal
equilibrium is achieved assuming no heat leaks?
(b) What is the entropy change of just the ice only while it
melts?
The latent heat of fusion and specific heats of ice and water are
3.34x105 J/kg, 2,108 J/(kgoC), and 4,184
J/(kgoC), respectively.
(Show all work
a) As you know the loss of the heat is equal to the gain of the heat
Here m=0.10 kg , S=4184 J/kgoC , M=0.40 kg , T2=30oC , T1=0oC , L=3.34x105 J/kg,
put these value in equation 1
0.40*4184*(30-T)=0.10*3.34x105 +0.10*4184 *(T-0)
50208 +1673.6 T =33400+418.4 T
1673.6T-418.4T=50208-33400
1255.2T=16808
T=16808/1255.2
T=13.39oC answer
b) We know that
S=mL/T1 +msln(T2/T1)
here T1=273.15 K , T2 =13.39oC=13.39+273.15 =286.54 K
S=0.1*3.34*10^5/273.15 +0.1*4184*ln(286.54/273.15)
S=122.28 J/K +20.02 j/K
S=142.3 J/K
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