A 51 g ice cube at 75degreeC is placed in a lake whose tempe

A 51 g ice cube at -75degreeC is placed in a lake whose temperature is 82degreeC. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. The specific heat of ice is 2220 J/kg-K.

Solution

The lake can be considered to be an infinite source of heat

So, heat absorbed by ice cube,

H = m*L + m*C_i*75 + m*C_w*82

where C_i = specific heat of ice = 2220 J/kg.K

C_w = specific heat of water = 4180 J/kg.K

L = latent heat of fusion of water = 334000 J/kg

So,

H = 0.051*2220*75 + 0.051*334000 + 0.051*4180*82

= 43006 J

So, change of entropy of the lake:

Sl = -H/T = -43007/(82+273) = -121.1 J/K

Entropy change for the ice:

Sw = 0.051*2220*ln(273/(273-75)) + 0.051*334000/273 + 0.051*4180*ln((273+82)/273)

= 154.8 J/K

So, the enntropy change of the system:

Snet = 154.8 - 121.1 = 33.7 J/K