Question

1.2 kg piece of ice at 0 oC is thrown into a very large lake at 12 oC. Water: Lf = 334 kJ/kg c = 4.19 kJ/(kg Co)

A) What is the entropy change of the ice? deltaS = ???J/K

B) What is the total entropy change of the ice and lake system? deltaS = ???J/K

I NEED NUMBERS PLEASE

Answer 1

Entropy change for ice:

entropy is changed in two phases in water -> 1. in melting at constant temperatue + 2 . in changing temperature from 0 to 12 C.

in melting Change in entropy = mL_f/T_MP = 1.2 x 334 /273 KJ/K = 1.4681 KJ/K

in increasing temp, Change in entropy  = mC ln(T₂/T₁) = 1.2 x 4.19 ln((12+273)/273) = 0.21629 KJ/K

so, change in entropy for ice = 1.4681 + 0.21629 = 1.68439 KJ/K

For lake: heat is removed at constant temperature of 12 C

change in entropy = Q/T

Q= -(mL_{f +}mC(T2-T1)) = -1.2 x (334+4.19x12)= -461.36 KJ

Q is negative because heat is rejected

T = 12 C = 12+273 = 285 K

so, Change in entropy of lake water = Q/T = -1.6180 KJ/K

for ice & lake system:

change in entropy = Entropy change for ice+Change in entropy of lake water = (1.68439-1.6180) KJ/K = 66.36 J/K