Respuesta :
Explanation:
The given data is as follows.
  [tex]m_{1}[/tex] = 30 kg,    [tex]m_{2}[/tex] = 40 kg
  [tex]T_{1} = (80 - 15)^{o}C[/tex] = [tex]65^{o}C[/tex]
  [tex]T_{2}[/tex] = [tex]65^{o}C[/tex]
Now, we will calculate the heat energy as follows.
    [tex]Q_{1} = m_{1}C \Delta T_{1}[/tex]
          = [tex]30 \times 0.45 \times 65^{o}C[/tex]
          = 877.5 kJ
and, Â Â [tex]Q_{2} = m_{2}C \Delta T_{2}[/tex]
           = [tex]40 \times 0.386 \times 65[/tex]
           = 1003.6 kJ
 [tex](\Delta S)_{lake} = \frac{(Q_{1} + Q_{2})}{T_{o}}[/tex]
        = [tex]\frac{(877.5 + 1003.6) kJ}{288}[/tex]
        = 6.531 kJ/K
 [tex]\Delta S_{1} = m_{1}C_{1} ln \frac{T_{2}}{T_{1}}[/tex]
       = [tex]30 \times 0.45 ln \frac{288}{353}[/tex]
       = -2.747 kJ/K
 [tex]\Delta S_{2} = m_{2}C_{2} ln \frac{T_{2}}{T_{1}}[/tex]
       = [tex]40 \times 0.386 \times ln \frac{288}{353}[/tex]
       = -3.142 kJ/K
Now, we will calculate the total entropy as follows.
  [tex]\Delta S_{total} = \Delta S_{lake} + \Delta S_{body}[/tex]
        = 6.531 kJ/K - 2.747 kJ/K - 3.142 kJ/K
        = 0.64 kJ/K
thus, we can conclude that the total entropy change for this process is 0.64 kJ/K.
