At a garden party on a hot summer day at a temperature of 30°C, somebody puts a

Question

At a garden party on a hot summer day at a temperature of 30°C, somebody puts a bowl of ice cubes (of total mass 0.50 kg) on a table for cooling drinks. After a while, 20% of the mass has melted. (a) Compute the entropy increase of the ice/water system due to the melting of the ice cubes and that due to the warming of the resulting water, within the assumption that the resulting (b) Explain why the entropy of the environment (air) also changes. Calculate the change in (c) Calculate the net change in the total entropy and compare the result to general expectations. water does not exchange an appreciable amount of heat with the remaining ice entropy of the air.

Explanation / Answer

T1=0 oC

T2=30oC

mass, m=0.5kg

percentage of mass melted 20%

a)

entopy, S1=m*L/T1 + m*C*ln(T2/T1)

=0.2*0.5*3.34*10^5/(273) + 0.2*0.5*4186*ln((273+3)/(273))

=126.92 J/K

b)

heat energy of the environment changes,

heat gain, Q=m*L_fusion + m*C_water*dT

Q=0.2*0.5*3.34*10^5 + 0.2*0.5*4186*(30-0)

Q=4.596*10^3 J

entropy, S2=Q/T2

=4.596*10^3/(303)

=15.17 J/K

c)

net change in entropy, Snet=S1+S2

=126.92+15.17

=142.09 J/K

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