Answer:
[tex]\delta \ S =0.034679 \ J/K[/tex]
Explanation:
The formula for calculating the change in entropy of a process can be expressed as:
[tex]\delta \ S = \int\limits \ \frac{d \theta }{T}[/tex]
[tex]\delta \ S = mc \int\limits^{T_2}_{T_1} \ \frac {d T}{T}}[/tex]
[tex]\delta \ S = mc \ In (T) ^{T_2}_{T_1}[/tex]
[tex]\delta \ S = mc \ [In \ (T_2) - In \ ({T_1})][/tex]
[tex]\delta \ S = mc \ In \ (\frac{T_2}{T_1})[/tex]
Given that:
mass m = 0.20 kg
specific heat constant c = 0.7186J/ Kg K
[tex]T_2[/tex] = 100° C = ( 100 + 273.15) = 373.15 K
[tex]T_1[/tex] = 20° C = ( 20 + 273.15) = 293.15 K
Replacing our values into above equation; we have :
[tex]\delta \ S = (0.20)(0.7186) \ In \ (\frac{373.15}{293.15})[/tex]
[tex]\delta \ S =0.034679 \ J/K[/tex]
