Respuesta :
Explanation:
Let us take the equilibrium temperature is T.
Now, the heat absorbed by the ice is equal to the heat lost to the water.
Therefore, the formula will be as follows.
[tex]m_{water} \times C_{water} (T - 87.3) + m_{ice} \times L_{f} + m_{ice} \times C_{water} (T - 0) = 0[/tex]
Therefore, putting the given values into the above formula as follows.
[tex]m_{water} \times C_{water} (T - 87.3) + m_{ice} \times L_{f} + m_{ice} \times C_{water} (T - 0) = 0[/tex]
[tex]150 g \times 4190 J/kg K (T - 87.3) + 0.0102 kg \times 333 kJ/kg + m_{ice} \times C_{water} (T - 0) = 0[/tex]
628500T - 54868050 + 3.3966 + 42.738T = 0
T = [tex]\frac{54868046.6}{628542.738}[/tex]
= [tex]87.29^{o}C[/tex]
or, = (87.29 + 273.15)K
= 360.44 K
Since, formula for entropy change of ice is as follows.
[tex]\Delta S_{ice} = \frac{m_{ice} \times L_{f}}{T_{ice}} + m_{ice} C ln \frac{T}{T_{melting}}[/tex]
= [tex]\frac{0.0102 [\frac{333000}{273.15} + 4190 \times ln (\frac{360.44}{273.15})][/tex]
= [tex]0.0102 \times 2381.015[/tex]
= 24.286 J/K
Entropy change for water ([tex]\Delta S_{water}[/tex]) = [tex]m_{ice} \times C \times ln (\frac{T}{T_{initial})}[/tex]
= [tex]0.0102 \times 4190 \times ln \frac{360.44}{360.45}[/tex]
= [tex]-1.186 \times 10^{-3}[/tex] J/K
Therefore, calculate the net change in entropy as follows.
[tex]\Delta S = \Delta S_{water} + \Delta S_{ice}[/tex]
= [tex]-1.186 \times 10^{-3} + 24.286[/tex] J/K
= 24.28481 J/K
Thus, we can conclude that the net entropy change of the system from then until the system reaches the final (equilibrium) temperature is 24.28481 J/K.
