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 the warming of the resulting water. (Look up the latent heat of fusion for water/ice.)

(b) Explain why the entropy of the environment (air) also changes. Calculate the change in entropy of the air.

(c) Calculate the net change in the total entropy and compare the result to general expectations.

Explanation / Answer

heat gain Q = m*L + m*c*dT = (0.2*0.5*334000)+(0.2*0.5*4186*30) = 45958 J

dS = mL/T + m*c*dT

S = mL/T +m*c*lnT

S = ((0.2*0.5*334000)/273) + 0.2*0.5*4186*ln(303) - ln273

S1 = 122.34 +43.64 = +165.98 J/K

b) air lose heat to sorroundings

S2 = 45958/303 = -151.67 J/k

c) S = S1+S2 = +14.31 J/K

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