Explanation:
According to the given situation, heat lost by coffee will be equal to heat gained by milk.
As we known that,
Q = [tex]mC \Delta T[/tex]
So, Heat energy lost by coffee = heat energy gained by milk
[tex]200 g \times C \times (70 - T) = 20 g \times C \times (T - 4)[/tex]
704 = 11T
T = [tex]64^{o}C[/tex]
As, [tex]\Delta S = \Delta S_{milk} + \Delta S_{coffee}[/tex]
[tex]\Delta S = 20 g \times C \times ln \frac{T_{f}}{T_{i}} + 200 g \times C \times n \frac{T_{f}}{T_{i}}[/tex]
= [tex]4.20 J/g^{o}C [50 ln \frac{(64 + 273)}{4 + 273} + 200 \times ln \frac{64 + 273}{70 + 273}][/tex]
= [tex]4.20 J/g^{o}C [50 \times 0.196 + 200 \times -0.0176][/tex]
= 26.376 J/K
Thus, we can conclude that the entropy increase of the Universe (in J/K) is 26.376 J/K.
