Question 2.54: 1 kg of ice at – 5°C is exposed to the atmosphere which is a...
1 kg of ice at – 5°C is exposed to the atmosphere which is at 27°C. The ice melts and comes into thermal equilibrium. Determine the entropy increase of the universe. For ice, c_p = 2.093 kJ/kg. °C. Latent heat of fusion of ice = 333.33 kJ/kg.
The system is shown in Fig 2.59.
\begin{aligned}&T_{\text {ice }}=-5+273=268 \mathrm{~K} \\&T_{\text {atm }}=27+273=300 \mathrm{~K}\end{aligned}
Heat absorbed by ice from atmosphere,
\begin{aligned}Q &=m L+m c_p \Delta t+m c_w \Delta t \\&=1 \times 333.33+1 \times 2.093 \times[0-(-5)]+1 \times 4.187 \times(27-0) \\&=333.33+10.465+113.049=456.844 \mathrm{~kJ}\end{aligned}
(a) Entropy change of atmosphere,
(\Delta S)_{\mathrm{atm}}=\frac{-Q}{T}=\frac{-456.844}{300}=-1.5228 \mathrm{~kJ} / \mathrm{K}
(b) Entropy change of ice from –5°C to 0°C,
(\Delta S)_1=\int\limits_{268}^{273} m c_p \frac{d T}{T}=1 \times 2.093 \ln \frac{273}{268}=0.0386 \mathrm{~kJ} / \mathrm{K}
Entropy change of ice from 0°C to water at 0°C,
(\Delta S)_2=\frac{L}{T_0}=\frac{333.33}{273}=1.2209 \mathrm{~kJ} / \mathrm{K}
Entropy change of water from 0°C to 27°C,
(\Delta S)_3=\int\limits_{273}^{300} m c_p \frac{d T}{T}=1 \times 4.187 \ln \frac{300}{273}=0.3948 \mathrm{~kJ} / \mathrm{K}
For ice :
\begin{aligned}(\Delta S)_{\text {Total }} &=(\Delta S)_1+(\Delta S)_2+(\Delta S)_3 \\&=0.0386+1.2209+0.3948=1.6543 \mathrm{~kJ} \mathrm{K}\end{aligned}
(c)
\begin{aligned}(\Delta S)_{\text {Universe }} &=(\Delta S)_{\text {Total }}+(\Delta S)_{\text {atm }} \\&=1.6543-1.5228=0.1315 \mathrm{~kJ} / \mathrm{K}\end{aligned}
