Question

1.10g of ice at 273K is added to 20 g of water at 90oC in an insulated flask. The heat of fusion of ice is 6 Kj/mol and the specific heat capacity of water is 4.2 J/K/g. ignoring the heat capacity of the flask;
Determine the final temperature of the system (3 marks)
Dtermine ∆S of the system (3 marks)

Answer 1

Let us write the energy conservation expression for this system:

$m_{wat.}c_{wat.}(t_{90}-t_{e})=m_{ice}\lambda_{ice}+m_{ice}c_{wat.}(t_{e}-t_{0});$

Thus,

$t_{e}=\frac{m_{wat.}c_{wat.}t_{90}-m_{ice}\lambda_{ice}+m_{ice}c_{wat.}t_{0}}{m_{ice}c_{wat.}+m_{wat.}c_{wat.}}=\frac{0,02kg*4200\frac{J}{kg*C}*90C-0.0011kg*330\frac{kJ}{kg}+0.0011kg*4200\frac{J}{kg*C}*0C}{0.0011kg*4200\frac{J}{kg*C}+0.02kg*4200\frac{J}{kg*C}}=81C;$

Entropy change could be expressed as:

$\delta S=S_{melting}+S_{heating}+S_{cooling};$

$S_{melting}=\frac{\delta H_{melting}}{T}=\frac{330*10^{5}\frac{J}{kg}}{273K}=1220\frac{J}{kg*K};$

$S_{heating}=S_{354}-S_{273}=C_{p}ln\frac{T_{2}}{T_{1}}=4200\frac{J}{kg*K}ln\frac{354K}{273K}=1091\frac{J}{kg*K};$

$S_{cooling}=S_{354}-S_{363}=4200\frac{J}{kg*K}ln\frac{354K}{363K}=-105\frac{J}{kg*K};$

Finally,

$\delta S=1220\frac{J}{kg*K}+1091\frac{J}{kg*K}-105\frac{J}{kg*K}=2206\frac{J}{kg*K}.$