Question

A small piece of ice (mass = 0.10 kg) at an initial temperature
of 0^{o}C is placed within glass of water, where the
water's mass is 0.40 kg and it is at an initial temperature of
30^{o}C.

(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
3.34x10^{5} J/kg, 2,108 J/(kg^{o}C), and 4,184
J/(kg^{o}C), respectively.

(Show all work

Answer #1

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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