Question

It can be shown that as a mass m with specific heat c changes temperature from Ti to Tf its change in entropy is ΔS=mcln(Tf/Ti) if the temperatures are expressed in kelvin. Suppose you put 78 g of milk at 278 K into an insulated cup containing 290 g of coffee at 355 K, and that each has the specific heat of water. The system comes to an equilibrium temperature of 339 K.

Part A

What is the entropy change of the milk?

Part B

What is the entropy change of the coffee?

Answer 1

part A

change in entropy of milk = S1 = Mmilk*Cmilk*ln(Tf/Ti)

Mmilk = mass of the milk = 0.078 kg

specific heat of milk = 4186 J/kg K

S1 = 0.078*4186*ln(339/278) = 64.77 J/K

part B

change in entropy of coffee = S2 = Mcoffee*Ccoffee*ln(Tf/Ti)

Mmilk = mass of the coffee = 0.29 kg

specific heat of coffee = 4186 J/kg K

S2 = 0.29*4186*ln(339/355) = -55.98 J/K